Signal Processing - An Introduction Processing – An Introduction ... “Analog and Digital Signal Processing”, ... – Biological: EEG interfered by ECG – Signal analytical:

Signal Processing

An Introduction

Magnus Danielsen

NVDRit 2007:13

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Heiti / Title Signal Processing An Introduction

Høvundar / Authors Magnus Danielsen

Ritslag / Report Type Undirvísingartilfar/Teaching Material

NVDRit 2007:13 © Náttúruvísindadeildin og høvundurin

ISSN 1601-9741 Útgevari / Publisher Náttúruvísindadeildin, Fróðskaparsetur Føroya Bústaður / Address Nóatún 3, FO 100 Tórshavn, Føroyar (Faroe Islands) Postrúm / P.O. box 2109, FO 165 Argir, Føroyar (Faroe Islands)

@ +298 352550 +298 352551 nvd@setur.fo

Abstract: English

Signal Processing – An Introduction Fundamentals of signal processing and analysis is treated on the base of Fourier representations, Laplace transformation, and z-transformation. Continuous and discrete Fourier representations are defined, and treated in depth with emphasis on illustrating the connection, similarities and non-similarities between the four Fourier representations of 1) discrete time periodic, 2) continuous time periodic, 3) discrete time non-periodic, and 4) continuous time non-periodic signals. Discrete Fourier transform is introduced, and the fast Fourier transform algorithms are defined and illustrated. The Laplace transform and the z-transform are defined and treated in the light of being a kind of generalization of the continuous and discrete Fourier transformations respectively. On the base of the signal representations a number of applications are dealt with. Sampling and reconstruction of signals, frequency analysis, fundamental analogue and digital filter constructions are treated, and a number of specific application examples are lined out. Føroyskt: Signalfrøði – Ein inngangur Grundleggjandi signalfrøði – viðgerð og greining – er gjøgnumgingin við støði í Fourier røðum og transformum, Laplace-, og z-transformum. Fourier røð og transformatiónir eru lýst og viðgjørd við denti á sambandinum, líkheitini og ólíkheitini millum tær fýra Fourier umboðanirnar: 1) diskret tíð periodisk, 2) kontinuert tíð ikki periodisk, 3) diskret tíð periodisk, og 4) kontinuert tíð ikki periodisk signal. Diskret Fourier transform er tikið upp, og “Fast Fourier Transform” algoritmur eru lýstar. Laplace transform and the z-transform hættirnir eru lýstir og viðgjørdir í ljósinum av at vera eitt slag av algilding av kontinuertum og diskretum Fourier transformatiónum og røðum. Við støði í hesum signal umboðanunum eru nýtslur av hesum hættum tiknar upp. Sampling og endurskapan av signalum er viðgjørd, frekvensgreining, grundleggjandi analog and digital filtur eru viðgjørd, umframt ein røð av serstøkum nýtsluendamálum eru greinað.

Contents: 1 Introduction

1.1 Introduction to course 1.2 Elementary signals and basic operations

2 Linear time-invariant systems 2.1 Linear time invariant systems and convolution 2.2 Properties of impulse response of a linear system 2.3 Sinusoidal response, diff. equations, and block diagrams

3 Continuous and discrete time Fourier representation of periodic and non-periodic signals 3.1 Fourier representations of signals 3.2 Discrete time (DTFS) and continuous time (FS) Fourier series 3.3 Discrete time non – periodic signals 3.4 Continuous time nonperiodic signals 3.5 Basic properties of Fourier Representations (1) 3.6 Basic properties of Fourier Representations (2) 3.7 Fundamental Operations and System Properties of Fourier Representations 3.8 Technical Applications of Fourier Representations

4 Sampling and reconstruction of signals 4.1 Sampling of signals 4.2 Aliasing and reconstruction of sampled signals

5 Discrete Fourier Transform and Fast Fourier Transform 5.1 Discrete Fourier Transform (DTF) for numerical evaluation 5.2 Fast Fourier Transform (FFT) – an efficient algorithm for numerical evaluation of DFT

6 Laplace Transform representation og signals 6.1 Laplace transform principles 6.2 The Laplace transform – transfer functions

7 Z-transform representation of signals 7.1 Digital filters – overview, and z-transforms 7.2 The z-transform and its properties 7.3 The z-transform – transfer functions 7.4 The z-transform – transfer functions. Frequency response and block diagrammes

8 Analogue and Digital Filters 8.1 Analogue filters 8.2 Analogue Butterworth filters 8.3 Analogue Chebyshev filters and frequency transformations of filters 8.4 Analogue filters – practical constructions 8.5 Digital FIR filter 8.6 Digital IIR filter – Low-pass, and high-pass filters 8.7 Digital IIR filter – Bandpass, and bandstop filters

9 Applications of signal processing in systems – some examples 9.1 Acoustic spectrogramme example 9.2 Delays in systems 9.3 Seismic example

Signal ProcessingPart 1.1 Introduction to Course

Magnus Danielsen

Textbooks incl. Supplementary Books

• S.Haykin. B.V.Veen: ”Signal and systems”, John Wiley and Sons, ISBN:0-471-13820-7 (Course book)

• M.Danielsen: “Phase delay and group delay” NVDRit 2007:02, (Course note)

• J.W. Nilsson, S.A. Riedel: “Electric Circuits”, 7th Edition, (background knowledge)

• H. Baher. “Analog and Digital Signal Processing”, 2nd Edition, John Wiley and Sons, ISBN: 0471-62354-7 (supplementary)

• P.A. Lynn, W. Fuerst, Introductory Digital Signal Processing, 2nd Edition John Wiley and Sons, ISBN: 0-471-97631-8 (supplementary)

• E.W.Kamen, B.S.Heck:”Fundamentals of Signals and Systems”, Prentice Hall, 2nd Edition, ISBN 0-13-017293-6 (supplementary)

• J.H.McClellan, “Signal Processing First”, ISBN 0-13-1201265-0 (supplementary)

• J.G.Proakis, D.G.Manolakis, “Digital Signal Processing”, Prentice Hall, ISBN 0-13-373762-4 (Advanced for electronics)

• B.Burrkus, ”Spectral and Filter Theory”, Springer, ISBN 3-540-62674-3 (Advanced for geophysics)

Contents of Course (1)• Introduction – basic signals and operations –

application examples• Time-domain – linear time invariant systems• Energy signals and power signals• Fourier transforms:

– Discrete time periodic signals DTFS– Continuous time periodic signals CTFS = FS– Discrete time non-periodic signals DTFT– Continous time non-periodic signals CTFT = FT

• Discrete Fourier transform and Fast fourier transform DFT & FFT

• Convolution/deconvolution• Sampling and reconstruction of signals

Contents of Course (2)• Laplace transform and its relations to Fourier

transforms• Z-transform and its relations to Laplace transform

and Fourier transforms• Analogue Filters • Digital FIR filters• Digital IIR filters• Equalizers• Application examples:

– Seismic geophone signal– Communication modulation wave forms– Speech signal spectrograms – Control feedback system

• MATLAB as signal processing tool

Types of Signals -Phenomenologically:

• Analogue signals– Function f(t) – Continous magnitude range– Continous argument t

• Discrete signals– Function f(t)=f(nT) or f(n) – Continuous magnitude range – Discrete argument t=nT or n (n is an integer)

• Digital signals– Function f(t) or f(n)– Discrete magnitude range– Discrete argument t=nT or n (n is an integer)

Types of Signals – Physically:• Sound

– Speech (100-10000 Hz)– Music (20-20000 Hz) – Telephone(300-3400 Hz)– Seismics (1-1000 Hz)– Ultra sound (20-100 kHz →)

• Light– Communications (glass fibres)– Infrared temp.measurements

• Electrical current and voltage• Sea-waves and -currents • Communications systems

(data, speech, pictures)• Internet & Email• Radiowaves

– Radar– Broadcasting (sound, pictures)– Communication

• Biological signals – ECG (hearth signals)– EEG (brain signals)– Blood pressure– Temperature

• Weather – Temperature– Moisture– Windspeed and direction

• Visual signals– Television– Computer display

• Statistics– World market– Finance, price, stock– Demography

• Etc.

Signal Processing System

• Communication system• Control system• Remote sensing system• Biomedical system• Auditory system• Audio system• Seismic system

SYSTEMInput signal Output signal

Communication System

• Message: Speech, TV/video, data• System: Digital or analogue• Channel: Opt.fibre, coax, radiolink, satellite,

mobile tel./PC• Disturbance: Noise, distortion, interference• Transceiv.: Coding, compression reconstruction,

modulation (AM, FM, PM, ASK, FSK, PSK, QAM etc.)• Electronics: VLSI• Modes: Broadcasting, point-to-point

Transmitter Channel ReceiverMessage signal

Transmitted signal

Received signal

Estimated received message signal

Disturbance

Control System

• Examples: – aircraft, autopilots, automobil engine, machine tools, oil raffineries,

oildrilling, papermills, fish industry, electrical powerplants, nuclear reactors, robots

• Response: Output follows input reference, regulation• Robustness: Good regulation despite of disturbances• Closed-loop: Feed-back control system• SISO: Single input / single output• MIMO: Multiple input / multiple output• Analogue systems: Analogue electronics• Digital systems: Digital electronics using microprocessors

Controller

Sensor

Plant ++Reference

Disturbance

Output+

Remote Sensing System

• Examples of fields:– Acoustic (sound, seismic, ultrasound), electromagnetic, electric, magnetic, gravitational– Passive (e.g. Earth quake), active (seismic)

• Sensors and systems:– Radar: surface properties, topography, roughness, moisture, dielectric constant– Infrared: near surface termal properties– Visual/near infrared: Chemical composition– X- and γ-ray: radioactive properties– Seismics and seismology: oil and gass exploration, earth quakes

• Digital signal processing - examples– Synthetic aperture radar (SAR): Use of single antenna in place of array, and use of FFT– Ocean seismic tecnology using airgun/hydrophone streamer

Passive system:

Sensor Signal processing

Input Signal

Object

Detected Signal

Processed Signal

Disturbance

Active system:

Sensor Signal processing

Received Signal

Object

Detected Signal

Processed Signal

Trans-mitter

Transmitted Signal Signal

modulator

Disturbance

Biomedical System

• Extraction of information from biological signals• Biological signals trace back to electrical activity• Neurons • ECG (EKG)• EEG• Artifacts

– Instrumental : thermal noise, 50Hz powerline noise – Biological: EEG interfered by ECG– Signal analytical: roundoff, quantization

• Signal processing: – Filtering– Use of apriori known signal properties: frequency range, randomlike properties,etc.

EEG: ECG:

Auditory System

Audio System - Spectrogram

Seismic System

A. Ziolkowski et al: “The signature of an air gun array: Computation from nearfield measurements including interactions”,Geophysics,vol. 47, oct. 1982, P. 1413

Signal ProcessingPart 1.2 Elementary Signals and Basic Operations

Magnus Danielsen

Contents

• Types of signals• Classification• Basic operations• Elementary signals• Systems, response and properties• Operators and block diagrams

Limitation of Signals Applied in this Course

• One dimensional signals and variables• Single valued signals• Real valued signals• Complex valued signals• Real independent variable (time or space)

Analogue signal = continuous signal:

Discrete signal:

Digital Signal

Analogue or Digital Signal Processing?

Analogue:– Real time– Simple circuits (often)– Physical in nature– Inflexible change of

system (rebuilding hardware)

– Suffers from parametrical variations

Digital:– Flexible– Software change of

system by changing programme

– Repeatibility– Complicated circuits

(VLSI)

Classification of Signals (1)1. Continuous / discrete time signals

– x(t)– x[n] with sampling time T

2. Even / odd signals– Even: x(-t)=x(t) x[-n]=x[n]– Odd: x(-t)=-x(t) x(-n)=-x[n]– Conjugate symmetric: x(-t)=x*(t) x[-n]=x*[n]

x(t)=a(t)+jb(t); a(-t)=a(t); b(-t)=-b(t) 3. Deterministic signals/

Indeterministic signals (random)

Classification of Signals (2)4. Periodic / non-periodic signals

– Periodic : x(t+T)=x(t) per.= T=1/f=(2π)/ωx[n+N]=x[n] per.= N=(2π)/Ω

– Non-per. : x(t) and x[n] not repeated periodically

5. Energy / power signals– Energy signals:

– Power signals:T/2 N

T N n NT/2

1 1lim x(t) dt lim x[n]T N→∞ →∞

=−−

< ∞ < ∞∑∫

x(t) dt x[n]∞ ∞

=−∞−∞

< ∞ < ∞∑∫

Basic Operations on Signals• Amplitude scaling

– y(t)=cx(t) y[n]=cx[n]• Addition

– y(t)=x1(t)+x2(t) y[n]=x1[n]+x2[n]• Multiplication

– y(t)=x1(t)·x2(t) y[n]=x1[n]·x2[n]• Differentiation / differensing

– y(t)=dx(t)/dt y[n]= x[n+1]-x[n]• Integration / summation

– y(t)=∫x(t)dt y[n]=Σx[n]

Basic Operation on the Independent Variable(1)

• Scaling– y(t)=x(at)

– Y[n]=x[kn]k = integer

• Reflection– y(t)=x(-t)

– y[n]=x[-n]

x[n] y[n]

Basic Operation on the Independent Variable

• Time shifting– y(t)=x(t-t0) – y[n]=x[n-m]

• Precedence operation– y(t)=x(at-b)– y[n]=x[pn-q]

Elementary Signals(1)• Exponential signal

– x(t)=B exp(at) a>0 growing signala<0 decaying signal

– x[n]=B rn r>1 growing; 0<r<1 decaying(B>0) r>0 positive x; r<0 alternating x

• Sinussoidal signal– x(t)=A cos(ωt+ϕ); x(t+T)=x(t); T=1/f=2π/ω

x[n]=A cos(Ωn+ϕ); – x[n+N]=x(t); Ω(n+N)+ ϕ = Ωn+2πm + ϕ

Periodicity requires: N=2πm/ Ω

Elementary Signals(2)

• Complex exponential and sinusoidal fct.– Eulers identities: ejθ = cos θ+jsin θ

cos θ = 1/2 (ejθ + ejθ)sin θ = 1/(2j) (ejθ - e-jθ)

– B = A ejθ

– A cos(ωt+θ) = ReB ejωt– A sin(ωt+θ) = ImB ejωt

Elementary Signals(3)

• Damped oscillation– Analogue: X(t) = A e-αt cos(ωt+θ) α > 0– Discrete: X[n] = B rn cos(Ωn+θ) 0 < r <1

• Step function– Analogue: u(t) = 0 t < 0

= 1 t ≥ 0– Discrete: u[n] = 0 t < 0

u[n] = 1 t ≥ 0 • Rectangular pulse

– p(t) = u(t + ½T) – u(t – ½T) p(t) = u(t) – u(t – T)– p[n] = u[n] – u[n – N]

Elementary Signals(4)• Impulse:

– Analogue:

– Discrete:

• Ramp: r(t) = t u(t) r[n] = n u[n]

0 t 0( t )

( t )dt 1 x ( t ) ( t t )dt x ( t )∞ ∞

−∞ −∞

≠δ = ∞ =

δ = δ − =∫ ∫

0 n 0( n )

1 n 0≠

δ = =

(t)δt

[n]δn

System and Operators Operator formulation:

Shift operator: x[n-k] = Skx[n] H = Sk (or St0)Differencing operator: y[n] = x[n-1] – x[n]

H = Sk – 1

Moving average operator: y[n] = 1/3·(x[n]+x[n-1]+x[n-2])

H = 1/3·(1+Sk+Sk 2)

Block diagram:

x[n] y[n] = Hx[n]

y(t) = Hx(t)

1/3y[n]

• StabilityBIBO: x ≤ Mx < ∞ ⇒ y ≤ My < ∞

• Memory: y(t0) depends on x for t<t0– memory: y[n] = 1/3·(x[n]+x[n-1]+x[n-2]) (Moving average )

i(t)=1/L ∫v(t)dt (Inductance)

– no memory: y[n]=x[n]2 (Square function)

v(t)=R i(t) (Ohms law)

• Causality:– y[n] depends only on x[m] for m ≤ n– y(t0) depends only on x(t) for t ≤ t0

Causal signal: y[n] = 1/3·(x[n]+x[n-1]+x[n-2])Noncausal sign.: y[n] = 1/3·(x[n+1]+x[n]+x[n-1])

Properties of System (1)Hx y

Properties of System (2)

Hxy=Hx

z=H-1H x=x

Invertibility – inverse system:– H-1H=I– H-1=inverse operator; I=identity operator

• Examples– Equalizer– Seismic signal detection– y(t)=1/L · ∫x(t)dt H= 1/L · ∫ H-1 =L·d/dt (inductance)– y=x2 x= ±y½ two solutions ⇒ non invertible

Properties of System (3)• Time invariance

Hxy=Hx

z= SkH x

Skxy=Hx

Hz=HSk x

Commutative rule: SkH = HSk

Properties of System (4)

• Linearity

i i i i

Input : x a x

Output : y H x a y a H x

∑ ∑

• Examples x= Σ ai xi– Linearity: y = nx = nΣ aixi = Σainxi = Σaiyi

– Non-linearity y = x(t)x(t-1) y = Σ ai xi(t)Σ aj xj(t-1) = Σ Σ ai aj xi(t) xj(t-1) ≠ Σ ai yi

Signal ProcessingPart 2.1 Linear Time Invariant Systems and Convolution

Magnus Danielsen

Contents of Part 2.1-2.3 • System• Impulse responce h(t) and h[n]• Convolution formulation for output: y=h∗x• Differential equations for analogue signals• Difference equations for discrete signals• Block diagram

– Scalar multiplication– Addition– Integration for analogue signals– Timeshift for discrete signals

• State variable description: – 1st order differential for analogue signals– 1st order difference equations for discrete signals

Impulse Response

Hx(t)=δ(t) y(t)=Hδ(t)=h(t)

Analogue signals

Hx[n]=δ[n] y[n]=Hδ[n]=h[n]

Discrete signals

x(t) y(t)t

x[n] n

Convolution for Discrete Signals

• Single pulse input: – Input signal: x[n]=x[k]·δ[n-k]– Output signal: y[n]=Hx[n]=h[n-k]

• Multi pulse input:– Input signal:

– Output signal:

x[n]= x[k] · [n-k]∞

k=- k=-

y[n]= x[k] · h[n-k]= x[n-k] · h[k] x[n] h[n]∞ ∞

∞ ∞

= ∗∑ ∑

Convolution for Analogue Signals

• Infiniticimal pulse input: – Input signal: dx=x(τ) · δ(t-τ)dτ– Output signal: dy=Hx(τ) · δ(t-τ)dτ=x(τ)h(t-τ)dτ

• Analogue function input:– Input signal:

– Output signal:

y(t) = x( ) · h(t- )d = x(t- ) · h( )d x(t) h(t)∞ ∞

∞ ∞

τ τ τ τ τ τ = ∗∫ ∫

x(t)= x( ) · (t- )d∞

τ δ τ τ∫

Discrete Convolution Graphically (1)x[n]=[-1 ½ 1 ½ -½] h[n]=[½ 1 ½ -½ ½ ¼ ]

vk[n]=pk[n] h[n-k]=x[k] h[n-k]nk=-1

y[n] p [n]h[n k]

x[k]h[n k]

=−∞

pk[n]=x[k]δ[n-k]

n-4 -3 -2 -1 0 1 2 3 4 5 6 7 8

Discrete Convolution Graphically (2)x[n]=[-1 ½ 1 ½ -½] h[n]=[½ 1 ½ -½ ½ ¼ ]

h[-1-k]

h[1-k]

h[2-k]

h[3-k]

h[4-k]

h[5-k]

h[6-k]

h[7-k]

h[8-k]

h[9-k]

y(n) x[k] h[n k]∞

−∞

= ⋅ −∑n=2

Discrete Convolution Calculation1 1 1 1 1h[n] [ 1 ]2 2 2 2 4

= −1 1 1x[n] [ 1 1 ]2 2 2

= − −

0½h(-2-k)

......

0...¼ h(9-k)

+1/8 = 0.125...½¼ h(8-k)

+1/8-¼ = - 0.125 ...-½½¼ h(7-k)

¼+¼+¼ = 0.75 ...½-½½¼ h(6-k)

+1/8+½-¼-¼ = 0.125 ...1½-½½¼ h(5-k)

-¼+¼-½+¼-½ = 0.75...½1½-½½¼ h(4-k)

-½-¼+½+½-¼ = 0...½1½-½½¼ h(3-k)

½+¼+1+¼ = 2 ...½1½-½½h(2-k)

-½+½+½ = 0.5...½1½-½h(1-k)

-1+¼ = - 0.75 ...½1½h(-k)

-½ = - 0.5 ...½1h(-1-k)

...0-½½1½-10...h(n-k) x(k)

543210-1-2...k=

y[n] x[k]h[n k]∞

=−∞

= −∑

Example on Discrete Convolution

3 2 13 3 34 4 4y[3] ( ) ( ) ( ) 1 2.734= + + + =

.... k

.... kh[3 k]−

.... kh[5 k]−

n 1n 3k 43

4 30 4

1 ( )y[n] 1 ( )1

+−= ⋅ =

−∑

10 1n 310 43

4 30 4

1 ( )y[10] ( ) 3.8311

+−= = =

−∑

x[k] u[k]=

y[ 5] 0− =

5 1n 3k 43

4 30 4

1 ( )y[5] ( ) 3.2881

+−= = =

−∑

( )n 1 2 33 3 3 34 4 4 4x[n] u[n] h[n] u[n] [..0 0 1 ( ) ( ) ( ) ....]= = =

Convolutional Integral for Analogue Signals (1)

Hx(t) y(t)x[n] y[n]

x[n]=δ[n-k] ⇒ y[n]=h[n-k]

x[n]=Σx[k]δ[n-k] ⇒ y[n]=Σx[k]·h[n-k]

x(t)=δ(t-τ) ⇒ y(t)=h(t-τ)= Hδ(t-τ)

y(t)=x(t)∗h(t)=h(t)∗x(t)

x( ) (t- )d y(t)=H x( ) (t- )d

= x( )h(t- )d = x(t- )h( )d

∞ ∞

τ δ τ τ ⇒ τ δ τ τ

τ τ τ τ τ τ

∫ ∫

Convolutional Integral for Analogue Signals (1)Example:

x(t)=e-3t u(t) – u(t-2) h(t) = e-t u(t)

t t-3 -(t- ) -t -2 t 2t t 3t

-3 -(t- ) t 2 2 4

y(t) e e d e e d ½e [e 1] ½(e e ) 0 t 2

e e d ½e [e 1] ½(1 e ) t 2

τ τ τ − − − −

τ τ − − ⋅ −

<= ⋅ τ = τ = − = − ≤ ≤ ⋅ τ = − − = − >

∫ ∫

Convolution of Discrete Step FunctionsNon - delayed step functions :

[ ] [ ] [ ] [ ]

( ) [ ]

u n u n u k u n k

n 1 u n=

∗ = ⋅ −

= + ⋅

∑ [ ] [ ] [ ] [ ]

( ) [ ]

1 2 1 2k n

1 2 1 2

u n n u n n u k n u n n k

n n n 1 u n n n

− ∗ − = − ⋅ − −

= − − + ⋅ − −

[ ] [ ]u n u n∗

.... k....u[n k]−

k=nk=0

.... k....u[k]

.... k....2u[n n k]− −

k=n-n2k=0

1u[k n ]−.... k....

k=n1k=0

n=0 n=n1+n2

[ ] [ ]1 2u n n u n n− ∗ −

Delayed step functions :

Moving Average System

[ ] [ ]N 1

1y n x n kN

where N is the averaging window width

= −∑Definition :

[ ]x n [ ]y n[ ]h n

[ ] [ ] [ ] [ ]

[ ] [ ]

[ ] [ ] [ ] [ ]( )

[ ] [ ]( )

x n n y n h n

1h n n kN1 n n 1 n 2 ... n N 1N1 u n u n NN

= δ ⇒ =

= δ −

= δ + δ − + δ − + + δ − +

= − −

Impulse response :

Moving Average of Rectangular Pulse[ ] [ ] [ ]x n u n u n M= − −

Moving average of rectangular pulse of width M :

[ ] [ ] [ ] [ ] [ ]( ) [ ] [ ]( )

[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]( )

( ) [ ] ( ) [ ](( ) [ ] ( ) [ ])

1y n h n x n u n u n N u n u n MN

1 u n u n u n N u n u n u n M u n N u n MN1 n 1 u n n N 1 u n NN

n M 1 u n M n N M 1 u n N M

= ∗ = − − ∗ − − =

∗ − − ∗ − ∗ − + − ∗ − =

+ − − + −

− − + − + − − + − −

Rectangular pulse of width M :

Averaging window width N=4

Pulse width M=10

.... n

n=N-1 n=M-1 n=N+M-1

Convolution of Continuous Step Functions

( ) ( )

u t u t

u u t d

= τ ⋅ − τ τ

( ) ( )

1 2 1 2

u t t u t t

u t u t t d

(t t t ) u t t t

− ∗ −

= τ − ⋅ − − τ τ

= − − ⋅ − −

Non - delayed stepfunctions : Delayed stepfunctions :

u(t)∗u(t)

u(t-τ)

τ = tτ = 0

τ = 0τ

τ=t1τ = 0

u(t-τ)

τ = t-t2τ = 0

u(t-t1)∗u(t-t2)

t = t1+t2t = 0

Signal ProcessingPart 2.2 Properties of Impulse Response

of a Linearsystem

Magnus Danielsen

Contents of Part 2.2

• Properties of impulse response– Parallel and cascade connections – Memory– Causality– Stability– Inversion– Step response

Properties of Impulse Respons (1)

x(t) +y(t)=x∗h1 + x∗h2

=x∗(h1 + h2)

Parallel connection: h = h1 + h2

Cascade connection: h = h1 ∗ h2

h1 h2x(t) y(t)=z∗h1 = x∗h1∗h2z(t)=

Exampley = - x∗h4 + x∗(h1 + h2)∗h3

= x∗(-h4 + (h1 + h2)∗h3)x

1h (n) u(n)= 2h (n) u(n 2) u(n)= + −

3h (n) (n 2)= δ − n4h (n) u(n)= α ⋅

[ ] [ ][ ]

4 1 2 3

h n h (h h ) h

u(n) (u(n) u(n 2) u(n)) (n 2)u n u n

(1 ) u n

= − + + ∗

= −α ⋅ + + + − ∗δ −

= −α ⋅ +

= − α ⋅

• Memoryless system

• Causal system

y[n] h[n] x[n]= h[k]·x[n-k] c·x[n]

h[n] c [n]

= ∗ =

⇒ = ⋅ δ

y(t) h ( t ) x ( t ) = h ( )·x(t- )d cx ( t )

h ( t ) c ( t )

= ∗ τ τ τ =

⇒ = ⋅ δ

k = k =

k = - k = 0

y [n ] h [ n ] x [ n ]= h [k ] ·x [n -k ]= h [k ] ·x [n -k ]∞ ∞

= ∗ ∑ ∑

y(t) h ( t ) x ( t ) = h ( )·x(t- )d = h( )·x(t- )d∞ ∞

= ∗ τ τ τ τ τ τ∫ ∫

• Stable systems: Bounded In – Bounded Out: BIBOx yx[n] M y[n] M≤ < ∞ ⇒ ≤ < ∞

y[n]= h[k] · x[n-k]∞

∞∑

x yy[n] h[k] · x[n-k] M h[k] <M h[k] <≤ ≤ ⇒ ∞∑ ∑ ∑

x yx(t) M y(t) M≤ < ∞ ⇒ ≤ < ∞

y(n)= h( )·x(t- )d∞

−∞

τ τ τ∫

x yy(t) h( ) · x(t- ) d M h( ) d <M h( ) d <∞ ∞ ∞

−∞ −∞ −∞

≤ τ τ τ ≤ τ τ ⇒ τ τ ∞∫ ∫ ∫

Discrete

Analogue

Examples on Stability

h[n] a u[n 2]1For 1 a 1: h[n] a a a BIBO stable

For a 1: h[n] unstable

∞ ∞− −

−∞ −

−∞

− < < = = + + ⇒−

≥ = ∞ ⇒

∑ ∑

Discrete system

Analogue systemat

h(t) e u(t)

e1For a 0 (i.e. e 1) : h(t) dt dt BIBO stablea

For a 0 (i.e. e 1) : h(t) dt unstable

−∞ ∞

−∞

∞−

−∞

> < = = ⇒

≤ ≥ = ∞ ⇒

∫ ∫

• Invertible systems and deconvolution – Examples:

• Equalizer in telecommunication:

• Seismic exploration

h(t)x(t) y(t)=h(t)∗x(t)

h-1(t)y(t) x(t)=h-1(t)∗y(t)

Transmitterh(t)

Receiverh-1(t)

Channelx(t) y(t)= h-1(t) ∗ h(t) ∗ x(t) = x(t)

x[n] y[n]=h[n]∗x[n]

h[n] = y[n] ∗ x-1[n]= h[n] ∗ x[n] ∗ x-1[n]

Example on Invertible Systems

h[n]x[n] y[n]= x[n] +ax[n-1]

y[n]=h[n]∗x[n]=Σh[n-k]x[k]Impulse response: h[n] = [1 a]

h-1[n]∗h[n] = Σ h-1[n-k] h[k] = δ[n]

h-1[0]·1 + h-1[-1]·a = 1 ⇒ h-1[0] = 1

h-1[1]·1 + h-1[0]·a = 0 ⇒ h-1[1] = -a

h-1[2]·1 + h-1[1]·a = 0 ⇒ h-1[2] = (-a)2

Inverse impulse response: h-1[n] = (-a)n = [1 -a (-a)2 (-a)3 (-a)4... ]

Undesired echo on data:

Step Response

h[n]u[n] s[n]=h[n]∗u[n]=Σh[n-k]u[k]

s(t) = h(t)∗u(t)=∫h(t-τ)u(τ)dτu(t)

s[n ] h[n ] u[n ] h[k ] u[n k ] h[k ]

s( t ) h ( t ) u ( t ) h ( ) u ( t )d h ( )d

= −∞ = −∞

−∞ −∞

= ∗ = ⋅ − =

= ∗ = τ ⋅ − τ τ = τ τ

∑ ∑

∫ ∫

Step response found from impulse response:

h[n ] s[n ] s[n 1]dh ( t ) s( t )d t

= − −

Impulse response found from step response:

Examples on Step Response

dE RI V I C V(t)dt

d(t) V(t) V(t) RC 1dt

V(0 ) (t)dt 1

Impulse response : V(t) h(t) exp( t)u(t)

Step response : s(t) h(t)dt (1 exp( t)) u(t)

−∞

δ = τ + τ = =

= δ =

= = −

= = − − ⋅

Analogue signal:

Discrete signal:

Impulse response: h[n] = (-a) u[n] a 1

1 ( a)Step response : s[n] h[n] u[n] (-a) u[n]1 a

∞ ∞

−∞

− −= = ⋅ = ⋅

+∑ ∑

E=δ(t) +−

V=h(t)−

Properties of the Impulse Response

Impulse response

Step response

h [n] ∗ hinv [n] = δ[n]h(t) ∗ hinv(t) = δ(t)Invertible

Stable

h[n]=0 for n<0h(t)=0 for t<0Causal

h[n]=c⋅δ[n]h(t)=c⋅δ(t)Memoryless

Discrete systemAnalogue systemProperty \ System

h(t) dt∞

−∞< ∞∫ n

h[n]∞

=−∞< ∞∑

ks[n ] h[k ]

= −∞= ∑

ts ( t ) h ( )d

−∞= τ τ∫

dh ( t ) s( t )d t

= h[n ] s[n ] s[n 1]= − −

Signal ProcessingPart 2.3 Sinusoidal Response, Diff. Equations, and Block Diagrams

Magnus Danielsen

Contents for Part 2.3

Sinussoidal response– Analogue systems– Discrete systems

• Differential equations • Difference equations• Block diagrams• State variables

hx[n]=ejΩn y[n]=h[n]∗x[n]=Σh[k]x[n-k]=Σh[k] ejΩ(n-k)=ejΩn · Σh[k] e-jΩk

y(t) = h(t)∗x(t)=∫h(τ)x(t-τ)dt= ∫h(τ) ejω(t- τ)dt= ejωt · ∫h(τ) e-jωτdτx(t) =ejωt

Discrete frequency response: H(ejΩ)=Σh[k] e-jΩk =H· ejϕ

y[n] = H(ejΩ)ejnΩ = H·x[n]

Analogue frequency response: H(ejω)=∫h(τ) e-jωτdτ =H· ejϕ

y(t) = H(ejΩ)ejωt = H·x(t)

Sinusoisal Steady State Response

Example on Sinusoidal Response

x[n]= A cos(Ωn+θ) y[n]

x(t) = A cos(ωt+θ)

Discrete frequency response:

x[n] = A cos(Ωn+θ) = ½A (ej(Ωn+θ) + e-j(Ωn+θ))

H = H(ejΩ) =H(ejΩ)· ejϕ(exp(jΩ)) =H· ejϕ

y[n] = H(ejΩ)· ejϕ(exp(jΩ)) ·½A ej(Ωn+θ) + H(e-jΩ)· ejϕ(exp(-jΩ)) ½A e-j(Ωn+θ)

= H· A cos(Ωn+ θ + ϕ)

Analogue frequency response:

x(t) = A cos(ωt+θ) = ½A (ej(ωt+θ) + e-j(ωt+θ))

H(jω) =H(jω)· ejϕ(jω)

y(t) = H(jω)· ejϕ(jω) ·½A ej(ωt+θ) + H(-jω)· e-jϕ(jω) ½A e-j(ωt+θ)

=H· A cos(ωt+ θ + ϕ)

If h is reel ⇒ H(ejΩ)= H(e-jΩ)* og H(jω)= H(-jω)*

Example on Sinussoidal Response Obtained from Impulse Response

• h1[n] = ½(δ[n] + δ[n-1])H1(ejΩ) = ½(1 + ejΩ) = ejΩ cos(Ω/2)

• h2[n] = ½(δ[n] - δ[n-1])H2(e-jΩ) = ½(1 - e-jΩ) = j e-jΩ sin(Ω/2)

Ω-π π

Ω-π π-π

Ω-π π-π/2

Example: Analogue Sinusoidal ResponsetRC

1h(t) e u(t)RC

11 RCH( j ) e u(t)e d 1RC j

∞ τ− − ωτ

−∞

ω = τ =+ ω

1RCH( j )1RC

ω = + ω

angle(H) arctan RC= − ω

-5 0 50

-5 0 5-1.5

Example: Discrete Sinusoidal Responsen

j j n n j n

h[n] ( a) u[n]

H(e ) h[n]e ( a) e

1H(e )1 a e

∞ ∞Ω − Ω − Ω

=−∞ =

= = −

=+ ⋅

∑ ∑

1H(e )(1 (a cos ) (a sin )

a sinangle(H) arctan1 (a cos )

Ω =+ Ω + Ω

Ω= −

Differential Equationsk kN M

k kk kk 0 k 0

d da y(t) b x(t)dt dt= =

=∑ ∑

x(t) +−

Example 1 t

dy 1L Ry ydt xdt C

d y dy 1 dxL R ydt dt C dt

−∞

Example 2

d y dym f ky xdt dt

Difference EquationsN M

k kk 0 k 0

a y[n k] b x[n k]= =

− = −∑ ∑

y[n] + y[n-1] + ¼ y[n-2] = x[n] + x[n-1]

y[n] = - y[n-1] - ¼ y[n-2] + x[n] + x[n-1]

Example

k kk 1 k 0

y[n] a y[n k] b x[n k]= =

= − − + −∑ ∑

Recursive formula for y[n]

Example Using Recursive Formula

Difference equation:

y[n] = – ½ y[n – 1] + ¼ y[n – 2] + x[n] – x[n – 5]

Initial output values: y[–1] = 1 y[–2] = 2Input: x[n] = u[n]

h[n]y[n]x[n]

Output solution:y[-2] = 2 = 2.00y[-1] = 1 = 1.00y[0] = –½ ⋅ 1 + ¼ ⋅ 2 + 1 – 0 = 1.00y[1] = –½ ⋅ 1 + ¼ ⋅ 1 + 1 – 0 = 0.75y[2] = –½ ⋅ 0.75 + ¼ ⋅ 1 + 1 – 0 = 0.89y[3] = –½ ⋅ 0.89 + ¼ ⋅ 0.75 + 1 – 0 = 0.75 y[4] = –½ ⋅ 0.75 + ¼ ⋅ 0.89 + 1 – 0 = 0.85 y[5] = –½ ⋅ 0.85 + ¼ ⋅ 0.75 + 1 – 1 = – 0.24y[6] = ½ ⋅ 0.23 + ¼ ⋅ 0.85 + 1 – 1 = 0.33 y[7] = –½ ⋅ 0.33 – ¼ ⋅ 0.23 + 1 – 1 = – 0.22y[8] = ½ ⋅ 0.22 + ¼ ⋅ 0.33 + 1 – 1 = 0.15y[9] = –½ ⋅ 0.15 – ¼ ⋅ 0.22 + 1 – 1 = – 0.13y[10]= ½ ⋅ 0.13 + ¼ ⋅ 0.15 + 1 – 1 = 0.10

-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

n5 7 9

-3 -2 -1 0 1 2 3 4 6 8 10

Block Diagrams - Symbolscx y=cx

+x y=x+w

∫x(t) y(t)= ∫t-∞ x(t)dt

Sx[n] y[n]=x[n-1]

Skx[n] y[n]=x[n-k]

Σx y=x+w

Block Diagram – Discrete 2nd Order

y[n] = b0x[n] + b1x[n-1] + b2x[n-2] - a1y[n-1] – a2y[n-2]

Block Diagram – Analogue 2nd Order

y = b0∫∫x + b1∫x + b2x – a0∫∫y’ – a1∫y

a0y + a1y’ + a2y’’ = b0x + b1x’ + b2x’’

a0∫∫y + a1∫y + a2y = b0∫∫x + b1∫x + b2x

Example –Block diagram for system with given difference equation

y[n] + 1/2 ⋅ y[n-1] – 1/3 ⋅ y[n-2] = 4 ⋅ x[n] – 3 ⋅ x[n-1] + 2 ⋅ x[n-2]

State Variable Description

y[n]b0

1 1 1 2 2

1 1 2 2 1 1 2 2

q [n 1] a q [n] a q [n] x[n]q [n 1] q [n]y[n] x[n] a q [n] a q [n] b q [n] b q [n]

+ = − − ++ == − − + +

1 11 2

11 1 2 2

q [n 1] q [n]a a 1x[n]

q [n 1] q [n]1 0 0

q [n]y[n] b a b a 1 x[n]

q[n 1] A q[n] bx[n]

y c q Dx

+ − − = + +

= − − +

+ = ⋅ +

= ⋅ +

Signal ProcessingPart 3.1 Fourier Representations of Signals

Magnus Danielsen

Fourier Representations of Signals

• Inventor: Joseph Fourier 1768 – 1830• Theory includes: Any ”wellbehaved signal” can be

represented as a continous (integral) or discrete (summation) superposition of sinusoids

• Sinussoids can be complex (mostly applied) or reel. • Complex fourier representations can be transformed into

real representations and vice verse• Fourier methods have widespread applications in signal

systems like– Communication systems– Seismic exploration systems– Audio systems – etc.

Fourier Series using Real Sinusoids

0 n 0 n 0n 1 n 1

f (t) a a cosn t b sinn t∞ ∞

= + ω + ω∑ ∑T

1a f (t)dtT

= ∫ ( )T

2b f (t)sin n t dtT

= ω∫( )T

2a f (t)cos n t dtT

= ω∫

2 2n n n

d a barctan(a / b )

a d cosb d sin

= ϕ= ϕ

f(t) is a periodic time functions with period T

0 n 0 nn 1

f (t) a d cos(n t )∞

= + ω −ϕ∑

• Eulers equations:cos x = 1/2·(ejx + e-jx) sin x = 1/2j·(ejx – e-jx)ejx = cos x + j sin x e-jx = cos x – j sin x

Fourier Series using Complex Sinusoids

0jn tn

f (t) c e∞

=−∞

= ∑ 0

1c f (t)e dtT

− ω= ∫

( )0 0

c ac ½ a jb for n 0=

= − ≠ ( )n n n

b j c c

Definition of Eigenvalue Formulation

( ) ( )( )

Eigenvalue formulation for the operator :

t eigenfunction = eigenvalue

ψ = λ ⋅ψ

Eigenvalue formulation for the matrix :

e eigenvector= eigenvalue

⋅ = λ ⋅

Linear Time Invariant (LTI) Systems –Complex Sinusoids Eigenfunctions for Analogue Systems

k kj t j tk k k

x(t) a e y(t) x(t) a H(j )e∞ ∞

=−∞ =−∞

= ⇒ = = ω∑ ∑

For LTI systems applies :

= h( ) corresponding to convolution operation

: (t) = e selected as input signal

y(t) = (t) =

e = h( )

τ ∗

Impulse response :

System operator : H

Eigenfunction

Output signal : H

H j t j ( t- ) j t -j j t

e h( ) e d e h( )e d H( j )e

: H( j ) h( )e d

∞ ∞ω ω τ ω ωτ ω

∞ ∞

∞ωτ

= τ ⋅ τ = τ τ = ω

λ = ω = τ τ

∫ ∫

∫Eigenvalue

H y(t) = H x(t)x(t)General LTI system:

Hy(t) = H(jω) ejωtψ(t) = ejωt

Eigenpresentation of system:

Linear Time Invariant (LTI) Systems –Complex Sinusoids Eigenfunctions for Discrete Systems

k k kj n j j nk k

x[n] a e y[n] x[n] a H(e )e∞ ∞

Ω Ω Ω

=−∞ =−∞

= ⇒ = =∑ ∑

For LTI systems applies :

[ ][ ]

[ ][ ] [ ]

[ ] [ ]

j n j n j (

= h n corresponding to convolution operation

: n = e selected as input signal

y n = n =

e = h n e h k e

Ω Ω Ω

∗ = ⋅

Impulse response :

System operator : H

Eigenfunction

Output signal : H

n k) j n j k j j n

k=- k=-

e h k e H(e )e

: H(e ) h k e

∞ ∞− Ω − Ω Ω Ω

∞ ∞

∞Ω − Ω

= ⋅ =

λ = = ⋅

∑ ∑

∑Eigenvalue

H y[n]= H x[n]x[n]General LTI system:

Hy[n] = H(ejΩ)ejΩnψ[n] = ejΩn

Eigenpresentation of system:

Fourier Representation for 4 Signal Classes

DTFT =Discrete – Time Fourier Transform

DTFS (→ *DFT) =Discrete – Time Fourier Series

Discrete

FT = CTFT = Continuous – Time Fourier Transform

FS = CTFS =Continuous – Time Fourier Series

Continuous

NonperiodicPeriodicTime property

*DFT = Discrete Fourier Transform

[ ] [ ] [ ] [ ]0 0

N 1 N 1jk n jk n

k 0 n 0

1x n X k e X k x n eN

− −Ω − Ω

= =∑ ∑Discrete Time Fourier Series = DTFS (→ DFT) :

Tjn t jn t

n nn 0

1f (t) c e c f (t)e dtT

∞ω − ω

=−∞

= =∑ ∫

Continuous Time Fourier Series = FS = CTFS:

[ ] ( ) ( ) [ ]j j n j j n

1x n X e e d X e x n e2

π ∞Ω Ω Ω − Ω

=−∞−π

= Ω =π ∑∫

Discrete Time Fourier Transform = DTFT:

( ) ( ) ( ) ( )j t j t1x t X j e d X j x t e dt2

π πω − ω

−π −π

= ω ω ω =π ∫ ∫

Continous Time Fourier Transform = FT=CTFT:

Definitions of Fourier Representations

Orthogonality

jk nk 0

N 1jk n jm n j( k m ) n

km k mN N n 0

j( k m ) N

j( k m )

D iscre te func tions:

n e N 2

I e e e

N for k m1 e N k m0 for k m1 e

−Ω − Ω − Ω∗

− Ω

φ = Ω = π

= φ φ = =

= −= = = ⋅ δ − ≠−

∑ ∑ ∑

jk tk 0

T Tjk t jm t j( k m ) t

km k mT t 0 t 0

Tj( k m ) t

Continuous functions:

( t ) e T 2

I ( t ) ( t )dt e e dt e dt

T for k me T k m0 for k mj(k m )

ω − ω − ω∗

− ω

φ = ω = π

= φ φ = =

= = = = ⋅ δ − ≠− ω

∫ ∫ ∫

Discrete -Time Periodic Signalswith Period N, and DTFS

[ ] [ ]

[ ] [ ] [ ] 0 0

jk n jk N

2x n x n N cyclic frequencyN

approximate x n by x n A k e e 1Ω Ω

π= + Ω =

= =∑

[ ] [ ]

[ ] [ ] [ ] [ ] [ ]0 0

jk n jk n

1Mean Square Error : MSE x n x n 0 whenN

1A k X k x n e and x n X k eN

− Ω Ω

= − =

∑ ∑

0DTFS,x[n] X[k]Ω←→

Proof of DTFS[ ] [ ]

0jk N0

: x n x n N2Cyclic frequency e 1N

πΩ = =

Given periodic discrete function

[ ] [ ] 0jk n

x n X k e Ω= ∑Consequently :

[ ] [ ] [ ]

[ ] ( ) [ ] [ ] [ ]

N 1 N 1 N 1jk n jk n ' jk n

k 0 k 0 n ' 0N 1 N 1 N 1

jk n n '

n ' 0 k 0 n ' 0

1x n X k e x n ' e eN

1 1x n ' e x n ' N n n ' x nN N

− − −Ω − Ω Ω

− − −Ω −

= = ⋅ ⋅ δ − =

∑ ∑∑

∑ ∑ ∑

Proof :

[ ] [ ] 0

N 1jk n '

1X k x n ' eN

−− Ω

= ∑Define :

Definition of DFT and FFT is most often defined

as a modification of DTFS, where the factor 1/N is moved from X[k] to x[n], to be used for a finite number N of data:

Discret Fourier Transform = DFT

is defined as specificallyefficient algoritms used to calculate Fast Fourier Transform = FFT

[ ] [ ]0 00

N 1 N 1jk n jk nDFT,

k 0 n 0

1x[n] X k e X[k] x n eN

− −Ω − ΩΩ

= ←→ =∑ ∑

Continuous Periodic Signalswith Period T, and FS=CTFS( ) ( )

( ) ( ) [ ] 0

2x t x t T cyclic frequencyT

approximate x t by x t A k e∞

=−∞

π= + ω =

( ) ( )

[ ] [ ] ( ) ( ) ( )0 0

jk t jk t

1Mean Square Error : MSE x t x t dt 0 whenT

1A k X k x t e dt and x t X eT

∞− ω ω

=−∞

= − =

= = = ω

∑∫

( ) 0FS,x t X[k]ω←→

Proof of FS( ) ( )

0jk T0

: x t x t T2Cyclic frequency e 1T

πω = =

Given periodic continuous function

( ) [ ] [ ] ( )0 0

Tjk t jk t

1ˆx t X k e and X k x t e dtT

∞ω − ω

=−∞

= =∑ ∫Consequently :

[ ] ( ) [ ]

[ ] ( ) [ ] [ ] [ ]

T Tjk t jk ' t jk t

k '0 0T

j k ' k t

k ' k '0

1 1X k x t e dt X k ' e e dtT T

1 1X k ' e dt X k ' T k ' k X kT T

∞− ω ω − ω

=−∞

∞ ∞− ω

=−∞ =−∞

= ⋅ = ⋅ ⋅ δ − =

∑∫ ∫

∑ ∑∫

Proof :

[ ] ( ) [ ] 0jk t

X k such that x t X k e∞

=−∞

= ∑Define :

Signal ProcessingPart 3.2 Discrete Time (DTFS) and Continuous Time (FS) Fourier Series

Magnus Danielsen

Example on DTFS[ ] 0

2x n cos( n ) Period N 16 cyclic frequency8 16 8π π π

= + ϕ = Ω = =

[ ] [ ]

8j n j n jk nj j8 8 8

x n ½(e e e e ) X k e

½e for k 1X k ½e for k 1

0 for k 1, 1

π π π−ϕ − ϕ

− ϕ

= −= = ≠ −

-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18k

X[k]½

∠X[k]

-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18k

Example on DTFS for Periodic Square WavePeriod = N cyclic frequency =Ω0

[ ] [ ] [ ]0 0 0

n M n M 2Mjk n jk M jk (n M)

n M n M 0

0jk (2M 1)

jk M 0jk

1 1X k x n e e x n eN N

1 sin(k (M ½) for k 0, N, 2N,...N sin(½k )1 1 ee

N 1 e 2M 1 for k 0, N, 2N,...N

= + =− Ω Ω − Ω +

=− + =

− Ω +Ω

− Ω

Ω + ≠ ± ± Ω− = = − + ≠ ± ±

∑ ∑

··· ··· ···N-M

···N

···

Discrete Time Fourier Serieswith Harmonically Related Cosines – Derivation

[ ] [ ] [ ] [ ]0 0jk n jk n

Ω − Ω= =∑ ∑

[ ] [ ] [ ] [ ] [ ]

[ ][ ]

N N 12 2

jk n jk n jk n j n

N k 1k 12

Nx n X k e X 0 X k e X k e X e2

B k cos(k n)

NX k for k 0,2B kN2X k for k 1,2,.., 12

Ω Ω − Ω − π

==− +

= = + + − +

== = −

∑ ∑

[ ] [ ] 0Even symmetry : X k X k Periodicity :N 2 N is even= − Ω = π

Discrete Time Fourier Series with Harmonically Related Cosines

[ ] [ ] 0Even symmetry : X k X k Periodicity :N 2 N is even= − Ω = π

[ ] [ ] [ ] [ ]0 0jk n jk n

Ω − Ω= =∑ ∑

[ ] [ ] [ ][ ]

NX k for k 0,2x n B k cos(k n) B kN2X k for k 1,2,.., 12

== Ω = = −

[ ] [ ] 0Even symmetry : X k X k Periodicity :N 2 N is odd= − Ω = π

[ ] [ ] [ ][ ]

X k for k 0x n B k cos(k n) B k N 12X k for k 1,2,..,

== Ω = −

Discrete Time Fourier Series with Harmonically Related Sines

[ ] [ ] 0Odd symmetry : X k X k Periodicity :N 2 N is even= − − Ω = π

[ ] [ ] [ ] [ ]0 0jk n jk n

Ω − Ω= =∑ ∑

[ ] [ ] [ ] [ ]N 12

Nx n B k sin(k n) B k 2 jX k for k 1,2,.., 12

= Ω = = −∑

[ ] [ ] 0Odd symmetry : X k X k Periodicity :N 2 N is odd= − − Ω = π

[ ] [ ] [ ] [ ]N 12

N 1x n B k sin(k n) B k 2 jX k for k 1,2,..,2

−= Ω = =∑

Fourier Transform – FS

[ ] ( ) 0

1X k x t e dtT

− ω= ∫

( ) ( ) 02x t x t T cyclic frequencyTπ

= + ω =

( ) ( ) 0jk t

x t X k e∞

=−∞

( ) [ ]

[ ] [ ] [ ] [ ]

T Tjk t jm t jk t

j(m k) t

1 1Proof: x t e dt X m e e dtT T

1X m e dt X m m k X kT

∞− ω ω − ω

=−∞

∞ ∞− ω

=−∞ =−∞

= = δ − =

∑∫ ∫

∑ ∑∫

Example – Cosine Functionj j t j j t4 2 4 23 3x(t) 3cos( t ) e e e e

2 4 2 2

π π π π− −π π

= + = +

[ ]X k

k-1 10

[ ]X k∠

Example – Sine Functions

j3 j2 t j3 j2 t j3(2 t ) j3(2 t )

x(t) 2sin(2 t 3) sin(6 t)j jje e je e e e2 2

− π − π π − π

= π − + π

= − + − +

[ ]X k

k-1 10 2 3-23

[ ]X k∠

-1 102π

− −

Example: FS of Square Wavex(t)

t-Ts Ts T T+TsT-Ts-T -T+Ts-T-Ts

[ ] ( )s

sT0 s s s s

Tsin(k2 )1 sin(k T ) 2T sin( u) 2T TTX k x t dt sinc(u) u=2T k k T u T T−

πω π= = = = =

π π π∫

- 5 0 - 4 0 - 3 0 - 2 0 - 1 0 0 1 0 2 0 3 0 4 0 5 0- 0 . 0 4

- 0 . 0 2

0 . 0 2

0 . 0 4

0 . 0 6

0 . 0 8

0 . 1 2

0 . 1 4

FS Approximation of Square Wave Using Finite Number of Terms

Haykin et al.

RC Circuit with Square Wave Input+

x(t) +−

01/ RCH( j ) k

j 1/ RCω = ω = ω

[ ] [ ]0

Tsin(k2

1/RCY kjk 1/RC

1/RCjk 1/RC

kksin10

j2 k 10

= ⋅ω +

= ⋅+ π

x(t) ↔X[k] is a rectangular per.pulse train with amplitude = 1 RC=0.1 s Ts/T=1/4 ω0 = 2π

-1 -0 .5 0 0 .5 10

( )y t

Time (s)

-2 0 -1 5 -1 0 -5 0 5 1 0 1 5 2 0-3

[ ]( )a rg Y k

-2 0 -1 5 -1 0 -5 0 5 1 0 1 5 2 0-0 .2

[ ]Y k

Signal ProcessingPart 3.3 Discrete Time Non – Periodic Signals

Magnus Danielsen

Discrete-time Nonperiodic Signals (1)DTFT

[ ][ ] [ ] [ ]

[ ] [ ]M

Non-periodic signal: x n for n

Define a periodic signal: x n x n+p N x n in period M n MLength of period : N 2M 1 (p is integer)

x n lim x n→∞

− ∞ ≤ ≤ ∞

= ⋅ = − ≤ ≤

[ ] [ ]

[ ] [ ] [ ]

M Mjk n jk n

2DTFS for per. signal : x n X k e2M 1

1 1X k x n e x n e2M 1 2M 1

− Ω − Ω

− −

π= Ω =

= =+ +

∑ ∑

[ ]Non-periodic x n

[ ]Periodic x n − M M

0n..... .....

( ) [ ] [ ]0

Mjk nj j t

2D efine : k d2 M 1

X e lim x n e x n e∞

− ΩΩ − Ω

→ ∞− −∞

πΩ = Ω Ω = Ω =

= =∑ ∑

[ ] [ ] [ ]

( ) ( )

M M k MM M

jk n ' jk n

M k M n ' M

j j n j j n

x n lim x n lim X k e

1lim x n ' e e2M 1

d 1X e e X e e d2 2

→∞ →∞=−

− Ω Ω

→∞=− =−

π πΩ Ω Ω Ω

−π −π

Ω = = Ω π π

∑ ∑

∫ ∫

Discrete-time Nonperiodic Signals (2) DTFT

DTFT Representation

[ ] ( ) ( ) [ ]j j n j j n1x n X e e d X e x n e2

π ∞Ω Ω Ω − Ω

−∞−π

= Ω =π ∑∫

( )DTFT, jx[n] X eΩ Ω←→

x[n] is non periodic and discrete in n

X e is periodic and continuous inΩ

• −

• Ω

Example on Exponential Sequence (1)

[ ] [ ]

( ) [ ]

j n j n j nj

x n u n

1X e u n e ( e ) for 11 e

∞ ∞Ω − Ω − Ω

− Ω−∞

= α = α = α <− α∑ ∑

( )( )( ) ( )

j½ ½2 22 2

1 1X e1 2 cos1 cos sin

Ω = =α + − α Ω−α Ω + α Ω

( )j sinX e arctan1 cos

Ω α Ω ∠ = − − α Ω

( ) ( ) ( )jj X ej jX e X e eΩ∠Ω Ω=

Example on Exponential Sequence (2)

- 1 5 - 1 0 - 5 0 5 1 0 1 50

- 1 5 - 1 0 - 5 0 5 1 0 1 5- 0 . 8

- 0 . 6

- 0 . 4

- 0 . 2

2π-2π

2π-2π 0

1=α +1-2αcosΩ

( )jX e

sinarctan1 cos

α Ω = − − α Ω

α = 0.5

Example: Rectangular Pulse (1)

( )j ( 2 M 1 )

j MMj j n j n

s in ( 2 M 1)1 e 2e 0, 2 , 4 .....

X e 1 e 1 e sin2

2 M 1 0, 2 , 4 .....

− Ω +− Ω

Ω − Ω − Ω

Ω + − = Ω ≠ ± π ± πΩ= ⋅ = −

+ Ω = ± π ± π

M = 10

[ ]1 for n 1

x n0 for n 1

≤= >

Example: Rectangular Pulse (2)

-1 5 -1 0 -5 0 5 1 0 1 5-5

M = 10( )jX e Ω

2π2− π 0

Example: Square Frequency Function – Discrete Sinc Time Function

-2 -1.5 -1 -0.5 0 0.5 1 1.5 20

− W− W

( )j 1 for WX e

0 for WΩ Ω ≤

= Ω >[ ] ( )sin Wn W Wnx n sinc( )

π π π

-40 -30 -20 -10 0 10 20 30 40-0.04

DTFT for x[n]=δ[n] and x[n]=1

( )DTFT j j n

x[n] [n] X e [n]e 1∞

=−∞

= δ ←→ = δ =∑

( ) ( ) ( )DTFTj j n1 1X e x[n] e d2 2

πΩ Ω

= δ Ω ←→ = δ Ω Ω =π π∫

Example: Sinusoidal Spectrum for a Discrete Non-periodic Signal (1)

[ ] ( )

j n j2 j2 j n

j (2 n) j ( 2 n)

Discrete Time Fourier Transform : X(e ) 2cos2

1 1x n 2cos2 e d e e e d2 2

1 1 1e e2 j(2 n) j( 2 n)

1 for n 21 for n 20 for n 2

π πΩ Ω − Ω Ω

−π −π

πΩ + Ω − +

= Ω ⋅ Ω = + ⋅ Ωπ π

= + π + − +

= −= = + ≠ ±

∫ ∫

Example: Sinusoidal Spectrum for a Discrete Non-periodic Signal (2)

-5 0 50

1.5[ ]Discrete time function x n

-4 -3 -2 -1 0 1 2 3 40

jAngle : X(e )Ω∠

-4 -3 -2 -1 0 1 2 3 40

jMagnitude : X(e ) 2 cos2Ω = Ω

Signal ProcessingPart 3.4 Continuous – Time Nonperiodic Signals

Magnus Danielsen

Continuous-time Nonperiodic Signals (1)FT = CTFT

( ) ( )

( ) ( )T

Non-periodic signal: x t for tT TDefine a periodic signal: x t x t period t2 2

x t lim x t→∞

− ∞ ≤ ≤ ∞

= − ≤ ≤

( ) [ ]

[ ] ( ) ( )

T T2 2

jk t jk t

T T2 2

2FS for per. signal : x t X k eT

1 1X k x t e dt x t e dtT T

−∞

− ω − ω

− −

π= ω =

∫ ∫

( )N on-periodic x t

t..... .....

( )Periodic x t − T/2 T/2

0t..... .....

( ) ( ) ( )0

jk t j t

2D efine : k dT

X j lim x t e d t x t e d t∞

− ω − ω

→ ∞−∞−

πω = ω ω = ω =

ω = =∫ ∫

Discrete-time Nonperiodic Signals (2) DTFT

( ) ( ) [ ]

( ) ( )

T / 2jk t ' jk t

T k T / 2

j t j t

x t lim x t lim X k e

1lim x t ' e dt ' eT

d 1X j e X j e d2 2

→∞ →∞=−∞

∞− ω ω

→∞=−∞ −

∞ ∞ω ω

−∞ −∞

ω = ω = ω ω π π

∑ ∫

∫ ∫

FT Representation

[ ] ( )FT,x t X jω←→ ω

( )( )

x t is non periodic and continuous in t

X j is non periodic and continuous in

• −

• ω − ω

( ) ( ) ( ) ( )j t j t1x t X j e d X j x t e d t2

∞ ∞ω − ω

−∞ −∞

= ω ω ω =π ∫ ∫

Example on Exponential Decay

( ) ( )at j t 1X e u t e dta j

∞− − ω

−∞

ω = =+ ω∫

( )2 2

1X(a )

ω =+ ω

( )X j arctanaω

∠ ω = −

-1 0 1 2 3 4 50

( ) ( )atx t e u t−=

Time t

2-3 -2 -1 0 1 2 3

( )X ω

Cyclic frequency ω

-3 -2 -1 0 1 2 3-1.5

( )X j∠ ω

Cyclic frequency ω

Rectangular Pulse

-20 -15 -10 -5 0 5 10 15 20-0.5

Cyclic frequency ω

2Tπ 3

-3 -2 -1 0 1 2 30

Time t

( ) ( )TT

j t j t j t

1 2 TX j x(t)e dt e dt e sin T 2Tsincj

∞− ω − ω − ω

−∞ − −

ω ω = = = = ω = − ω ω π ∫ ∫

Rectangular SpectrumWπ

( ) ( )WW

j t j t j t

1 1 1 1 W Wtx(t) X j e dt e dt e sin Wt sinc2 2 j t

∞ω ω ω

−∞ − −

= ω = = = = π π ω π π π ∫ ∫

FT for an Impulse

FT for a DC signal

( ) ( )FT j tt t e dt 1∞

− ω

−∞

δ ←→ δ =∫

( ) ( ) ( )FT j t1 1x t e d2 2

−∞

δ ω ←→ = δ ω ω =π π∫

( )FT1 2←→ πδ ω

FT of Ramp Function (1)

[ ]t t 1

x(t) t u(t 1) u(t 1)0 t 1

≤= = ⋅ + − − >

( ) ( )

11 1j t j t j t

1 1112

j t j t

1 1X( ) t e dt t e e dtj j

1 1t e ej j

2 2j cos j sin

− ω − ω − ω

− −−

− ω − ω

ω = ⋅ = ⋅ − − ω − ω

= ⋅ − − ω − ω

= ω − ωω ω

∫ ∫

FT of Ramp Function (2)

-20 -15 -10 -5 0 5 10 15 20-100

0Magnitude:

Cyclic frequency ω

-20 -15 -10 -5 0 5 10 15 20-2

2Phase:

Cyclic frequency ω

( ) ( )2

X( )2 2j cos j sin

ω − ωω ω

Binary Phase Shift Keying (BPSK) Digital Communication Signal (1)

Single rectangular pulse shape:

Single raised cosine pulse shape:

½T2 2

r0 ½T

1Power : P x (t) dt AT −

= =∫

½T2 2

c0 ½T

1 3Power : P x (t) dt AT 8−

= =∫

t-½T0 ½T0

Axr(t)

-0 .5 0 0 .50

t-½T0 ½T0

Axc(t)

A t ½Tx (t)

0 t ½T≤

½A(1 cos(2 t / T )) t ½Tx (t)

0 t ½T+ π ≤

Raised cosine pulse shaped BPSK

0 2 4 6 8 10-1

1 0 1 1 0 1 0 0 1

Rectangular pulse shaped BPSK1 0 1 1 0 1 0 0 1

Binary Phase Shift Keying (BPSK) Digital Communication Signal (2)

-10 -5 0 5 10-150

0Rectangular pulse, P=1 ⇒ A=1:

Raised cosine pulse, P=1 ⇒ A=(8/3)½ :

Power spectrum:

20⋅lo

Rectangular pulse

Raised cosine pulse

Frequency f (unit = 1/T0)

P=1 ( )

T / 2j2 ft

rT / 2

X ' jf 1 e dt

sin fTf

− π

( )( )( )( )( )( )

T / 2j2 ft

c 0T / 2

1 8X ' jf (1 cos(2 t / T )e dt2 3

sin fT23 f

sin f 1/ T T20.53 f 1/ T

− π

= + π

ππ −

+ ⋅π −

π ++ ⋅

Signal ProcessingPart 3.5 Basic Properties of Fourier Representations (1)

Magnus Danielsen

Basic Properties of Fourier Representations

• 4 Fourier Representations• Periodicity• Linearity• Symmetry• Time-shift• Frequency-shift• Scaling• Differentiation/ differencing in time• Differentiation/ differencing in frequency• Integration/summation

Frequencyproperty

Continuous Discrete

Periodic

Non-periodic

DTFT =Discrete – Time Fourier Transform

DTFS = DFT =Discrete – Time Fourier Series

Discrete

FT = CTFT = Continuous – Time Fourier Transform

FS = CTFS = Continuous – Time Fourier Series

Conti-nuous

NonperiodicPeriodicTime property

[ ]0FS,x(t) X kω→ [ ]FTx(t) X j→ ω

[ ] [ ]0DTFS;x n X kΩ→ [ ] [ ]DTFTx n X k→

Linearityx X y Y↔ ↔

z ax by Z aX bY= + ↔ = +

[ ] 1Z k 3sin k sin k2k 4 2

π π = + π

[ ] 1Y k sin kk 4

π = π

[ ] 1X k sin kk 2

π = π

Example with Fourier series:

t-1/8 1/8 11

t1/4-1/4

z(t)=3/2·x(t)+1/2·y(t)

Linearity and Decomposition - FT

( ) ( )( )

( )( )

M NM kpkp

k 0p 0N

B jf j for M Nb j A jB j

X jA j B ja j for M N

ω⋅ ω + ≥ω ωω ω = = = ω ωω < ω

∑∑

( ) ( )( )

N Pk,pk

k pk 1 p 2 fork k

P mult.poles

DB j CFor M N f 0 X j ( )A j j d ( j d )= =

ω< ⇒ = ω = = +

ω ω− ω−∑ ∑

( ) ( ) ( )k

FT FT FT kk

d dt 1 t j t ( j )dt dt

δ ←→ δ ←→ ω δ ←→ ω

( ) ( )k kd t d tFT FTpp

1 (p 1)!x t e x t t ej d ( j d )

− − −= ←→ = ⋅ ←→

ω− ω−

Linearity and Decomposition - DTFT

( ) ( )( )

( )( )

jM N kM jpj k jj p k 0p 0jNj jqj

B ef e for M Nb e A eB e

X eA e B ea e for M N

− Ω−− Ω− Ω

− Ω− Ω==Ω

− Ω − Ω− Ω

− Ω=

⋅ + ≥

∑∑

( ) ( )( )

j N Pk,pj k

k j j pjk 1 p 2 fork k

P mult.poles

B e DCFor M N f 0 X e ( )1 d e (1 d e )A e

− Ω− Ω

− Ω − Ω− Ω= =

< ⇒ = = = +− −∑ ∑

[ ] [ ] [ ] 0j nDTFT DTFT DTFTj0n 1 n 1 e n n e− Ω− Ωδ ←→ δ − ←→ δ − ←→

[ ] [ ] ( )

DTFTn jj

p 1DTFTp n j

1x n d u n X e1 de(p 1)! ( d)x n n d u n X e(1 de )

Ω− Ω

− Ω

= ←→ =−

− ⋅ −= ←→ =

Examples on Decomposition

j 1X j( j ) 5j 6C C 1 2j 2 j 3 j 2 j 3

ω+ω =

ω + ω+−

= + = +ω+ ω+ ω+ ω+

( ) j2t j3tx t e 2e− −= − +FT←→

FT with 2 different poles

5 e 56X e 1 1(e ) e 1

6 6C C1 11 e 1 e2 3

4 11 11 e 1 e2 3

− Ω

− Ω − Ω

− +=− + +

= ++ −

= ++ − [ ] [ ] [ ]

n n1 1x n 4 u n u n2 3

= − +

DTFT←→

DTFT with 2 different poles

Symmetry for Real Valued Time Signals

( ) ( ) j tFT for non periodic signal x(t) : X j x t e dt∞

− ω

−∞

− ω = ∫

( ) [ ] ( ) 0jk t0

2 1FSfor per.signal x t , : X k x t e dtT T

− ωπω = = ∫

[ ] ( ) [ ]j j nDTFT for non periodic signal x n : X e x n e∞

Ω − Ω

−∞

− =∑

[ ] [ ]X k X k∗ = −

j jX e X e∗ Ω − Ω =

[ ] [ ]X j X j∗ ω = − ω

[ ] [ ] [ ] 0jk n0

2 1DTFS for periodic x n , : X k x n eN N

− ΩπΩ = = ∑

Symmetry for Imaginary Valued Time Signals

− ω

−∞

− ω = ∫

( ) [ ] ( ) 0jk t0

− ωπω = = ∫

Ω − Ω

−∞

− =∑

[ ] [ ]X k X k∗ = − −

j jX e X e∗ Ω − Ω = −

[ ] [ ]X j X j∗ ω = − − ω

[ ] [ ] [ ] 0jk n0

− ΩπΩ = = ∑

Even and Odd Real Valued Function Symmetries

− ω

−∞

− ω = ∫

( ) [ ] ( ) 0jk t0

− ωπω = = ∫

[ ] [ ] [ ] 0jk n0

− ΩπΩ = = ∑

Ω − Ω

−∞

− =∑

Even valued real functions x(-t)=x(t) and x[-n]= x[n]: X* = X ⇒ ImX=0

Odd valued real functions x(-t)=x(t) and x[-n]= -x[n]: X* = - X ⇒ ReX=0

Time - Shift

( ) ( )

j t0 t 0

j ( t t ) j t

FT x(t t ) : X j x t t e dt

x t e dt e X( j )

∞− ω

−∞

∞− ω + − ω

−∞

− ω = −

= = ω

( ) [ ] ( )

( ) [ ]

0 0 0 0

jk t0 t 0

jk ( t t ) jk t

1FS x t t : X k x t t e dtT

1 x t e dt e X kT

− ω

− ω + − ω

− = −

[ ] [ ] [ ]

[ ] [ ]

0 0 0 0

jk n0 n 0

jk ( n n ) jk n

1DTFS x n n : X k x n n eN

1 x n e e X kN

− Ω

− Ω + − Ω

− = −

[ ] ( ) [ ]

[ ] ( )

j j n0 n 0

j ( n n ) j n j

DTFT x n n : X e x n n e

x n e e X e

∞Ω − Ω

−∞

∞− Ω + − Ω Ω

−∞

− = −

∑ ( )0j nDTFT j0x[n n ] e X e− Ω Ω− ←→

[ ]0 00 jk nDTFS,0x[n n ] e X k− ΩΩ− ←→

( ) [ ]0 00 jk tFS,0x t t e X k− ωω− ←→

( ) ( )0j tFT0x t t e X j− ω− ←→ ω

Example on Time Shift

( ) FT 2x t sin( T)←→ ωω

( ) ( ) ( )FT j T

z(t) x t T Z j e X j2e sin( T)

− ω

= − ←→ ω = ω

= ωω

- 4 - 3 - 2 - 1 0 1 2 3 4- 5

- 4 - 3 - 2 - 1 0 1 2 3 40

- 4 - 3 - 2 - 1 0 1 2 3 4- 4

- 4 - 3 - 2 - 1 0 1 2 3 40

X(ω) Z(ω)= X(ω)

∠ Z(ω)∠ X(ω)

-π0Cyclic frequency ω Cyclic frequency ω

Signal ProcessingPart 3.6 Basic Properties of Fourier Representations (2)

Magnus Danielsen

Frequency Shift( )FTx(t) X j←→ ω

( ) ( )

j t j( ' ) t j t

Z j X j( )

1 1z(t) X j( ) e d X j ' e d ' e x(t)2 2

∞ ∞ω ω +γ γ

−∞ −∞

ω = ω− γ ←→

= ω− γ ω = ω ω =π π∫ ∫

[ ] [ ][ ]

[ ] ( )( )

jk n DTFS,0

jk t FS,0

DTFTj t j( )

e x n X k k

e x(t) X k k

e x n X e

e x(t) X j( )

Γ Ω−Γ

⋅ ←→ −

⋅ ←→

⋅ ←→ ω− γ

Example on Frequency Shift for FT

j10tj10te for t

z(t) e x(t)0 for t

≤ π = = > π

1 for tx(t)

0 for t≤ π

= > π

( ) ( ) ( )2X j sin 2 sincω = ωπ = π ωω

( ) ( )

2Z j sin ( 10)10

2 sinc 10

ω = ω− πω−

= π ω−

-3 -2 -1 0 1 2 30

Time t-π π

-1 5 -1 0 -5 0 5 1 0 1 5-2

X(jω) Z(jω)

Cyclic frequency ω

-3 -2 -1 0 1 2 3-1 .5

-π π

0Time t

Rez(t)

Example on Frequency Shift for DTFT

[ ] [ ] [ ]j n j nn4 4z n e u n e x nπ π

− −= α =

[ ] [ ] ( )DTFTn jj

1x n u n X e1 e

ΩΩ= α ←→ =

( ) ( )j j( / 4)j( / 4)

1Z e X e1 e

Ω Ω+πΩ+π= =

− α

Scaling

( )jFT j t a1 1z(t) x(at) Z j x(at)e dt x( )e d X j

∞ ∞ τ− ω− ω

−∞ −∞

ω = ←→ ω = = τ τ = ∫ ∫

Example on rectangular pulse: a=½

-3 -2 -1 0 1 2 30

Time t-1 1

( ) ( )2X j sinω = ωω

( ) ( )1 2 2Z j sin sin 2½ /½ ½

ω ω = = ω ω ω

- 3 - 2 - 1 0 1 2 30

Time t-2 2

z(t)=x(½t)

( )FTx(t) X j←→ ω

-20 -15 -10 -5 0 5 10 15 20-1

Scaling for Analogue Time Functions

( )FTx(t) X j←→ ω

( )jFT j t a1 1z(t) x(at) Z j x(at)e dt x( )e d X j

∞ ∞ τ− ω− ω

−∞ −∞

ω = ←→ ω = = τ τ = ∫ ∫

[ ] [ ]00 jka tFS,a

az(t) x(at) Z k x(at)e dt X kT

ωω= ←→ = =∫

a = any real number

[ ]0FS,x(t) X jω←→ ω

Continous Time Fourier Transform

Continous Time Fourier Series

Scaling for Discrete Time Signalsp = any integer number

[ ] [ ] [ ]

j j nz

j n jp p

z n x p n x p n

Z e x [pn]e

x [n]e X (e )

∞Ω − Ω

=−∞

Ω Ω∞ − −

=−∞

= ⋅ = ⋅ ←→

[ ] ( )DTFT jx n X e Ω←→

Discrete Time Fourier Transform

[ ] [ ] [ ] [ ] [ ]0DTFS,pz zz n x pn x pn Z k pX kΩ= = ←→ =

[ ] [ ]0DTFS,x n X kΩ←→Discrete Time Fourier Series

nxz [n]

Differentiation in Time

( ) ( ) ( ) ( )j t j t1x t X j e d X j x t e dt2

∞ ∞ω − ω

−∞ −∞

= ω ω ω =π ∫ ∫

( ) ( ) ( )F Tj td 1x t X j ( j ) e d ( j ) X jd t 2

−∞

= ω ⋅ ω ⋅ ω ←→ ω ωπ ∫

( ) ( )FTd x t ( j )X jdt

= ←→ ω ω

[ ] [ ]0 0

Tjk t jk t

1x(t) X k e X k x(t)e dtT

∞ω − ω

=−∞

= =∑ ∫

Continuous Time Fourier Series, FS

Continuous Time Fourier transform, FT

( ) [ ]0FS,0

d x t jk X kdt

ω= ←→ ω

Differencing in Time

[ ] [ ] [ ] [ ] [ ]n

kj j j j

y n x k x n y n y n 1

X(e ) Y(e ) e Y(e )=−∞

Ω Ω − Ω Ω

= = − −

[ ] DTFT j j jx n X(e ) (1 e )Y(e )Ω − Ω Ω←→ = −

Discrete Time Fourier transform, DTFT

[ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ]0

y n x k x n y n y n 1

X k Y k e Y k=−∞

− Ω

= = − −

[ ] [ ] [ ]00 jkDTFS,x n X k (1 e )Y k− ΩΩ←→ = −

Discrete Time Fourier Series, DTFS

Differentiation in Frequency

( ) ( ) ( ) ( )j t j t1x t X j e d X j x t e d t2

∞ ∞ω − ω

−∞ −∞

= ω ω ω =π ∫ ∫

( ) ( ) ( )FTj td X j x t ( jt)e dt jt x td

∞− ω

−∞

ω = − ←→ − ⋅ω ∫

( ) ( )FTd X j jt x td

ω ←→− ⋅ω

Continuous Time Fourier transform, FT

[ ] DTFT jdjn x n X(e )d

Ω− ⋅ ←→Ω

Discrete Time Fourier transform, DTFT

( ) [ ] ( ) [ ]j j n j j n

dX e x n e X e jn x n ed

∞ ∞Ω − Ω Ω − Ω

=−∞ =−∞

= = − ⋅Ω∑ ∑

Differencing in Frequency

[ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ]0

Y k X k X k Y k X k 1

x n y n e y n=−∞

= = − −

[ ] [ ] [ ] [ ] [ ]0 0jn DTFS,x n (1 e )y n X k Y k X k 1Ω Ω= − ←→ = − −

Discrete Time Fourier Series, DTFS

Differencing in Frequency

[ ] [ ]0 0

Tjk t jk t

1x( t) X k e X k x( t)e dtT

∞ω − ω

=−∞

= =∑ ∫

Continuous Time Fourier Series, FS

[ ] [ ] [ ] [ ] [ ]

( ) ( ) ( )0

Y k X m X k Y k Y k 1

x t y t e y t=−∞

= = − −

( ) [ ] [ ] [ ]0 0j t FS,x(t) (1 e )y t X k Y k Y k 1ω ω= − ←→ = − −

Integration

( ) ( ) ( ) ( ) ( )t

FT 1y(t) x d Y j X j X j0j

where first term equals zero for 0 −∞

= τ τ←→ ω = ω + π δ ωω

( ) ( ) ( ) ( )

( ) ( ) ( )

tt tj t j t j t

dy(t) x d y t x(t) j Y j X jdt

1 1Y j x d e dt x d e x(t)e dtj j

1X( j0) ( ) X( j )j

−∞

=∞∞− ω − ω − ω

−∞ −∞ −∞ =−∞

= τ τ = → ω ω = ω

ω = τ τ = τ τ ⋅ − − ω − ω

= ⋅ πδ ω + ωω

∫ ∫ ∫ ∫

Fourier Transform, FT

Example of Integration u(t) = ∫δ(t)dt

( ) ( )

Continuous impulse : t

Step function : u t t dt

1Fourier transform : U( j ) 1j

−∞

ω = + π ⋅ ⋅ δ ωω

Use of the integration formula :

( ) ( )

1 1Step function : u t sgn t2 2

2 0jFourier transform : sgn t0 0

1 1 22 2

1Fourier transform : U( j )j

ω = ω←→ ω =

←→ πδ ω = πδ ω

ω = + πδ ωω

Alternative calculation :

Summation[ ] [ ]

y n x k=−∞

[ ] ( )DTFT j j j0j

DTFT1y n Y(e ) X(e ) X(e )

(1 e )for

where first term equals zero for 0

Ω Ω− Ω←→ = + π δ Ω

−− π ≤ Ω ≤ π

[ ] [ ] [ ] DTFT j j jy n y n 1 x n X(e ) (1 e )Y(e )Ω − Ω Ω− − = ←→ = −

Signal ProcessingPart 3.7 Fundamental Operations and System Properties of

Fourier Representations

Magnus Danielsen

Fundamental Operations and System Properties of Fourier Representations

• Convolution• Modulation• Power spectrum and Parceval relationships• Duality• Time – bandwidth product• Auto – correlation • Cross – correlation

Non-periodic Convolution (FT)

( ) ( )

( ) ( ) ( )

( ) ( )

y( t) h( t) x(t) h x t d

Y j h x t d e dt

h x t e dt d

h X j e d

H j X j

−∞

∞ ∞− ω

−∞ −∞

∞ ∞− ω

−∞ −∞

∞− ωτ

−∞

= ∗ = τ − τ τ

ω = τ − τ τ

= τ − τ ⋅ τ

= τ ω τ

= ω ⋅ ω

∫ ∫

( )FTy(t) Y j←→ ω( )FTx(t) X j←→ ω( )FTh(t) H j←→ ω

Non-periodic Convolution (DTFT)

[ ] [ ] [ ] [ ] [ ]

( ) [ ] [ ]

[ ] [ ]

( ) ( )

j k j ( n k )

y n h n x n h k x n k

Y e h k x n k e

h k e x n k e

H e X e

=−∞

∞ ∞Ω − Ω

=−∞ =−∞

∞ ∞− Ω − Ω −

=−∞ =−∞

= ∗ = −

= − ⋅

∑ ∑

[ ] ( )DTFT jy n Y e Ω←→[ ] ( )DTFT jx n X e Ω←→

[ ] ( )DTFT jh n H e Ω←→

Periodic Convolution (DTFS)

[ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ]N

y n h n x n h k x n k

Y k N H k X k

= ∗ = ⋅ −

= ⋅ ⋅

[ ] [ ]02DTFS,Ny n Y kπ

Ω =←→[ ] [ ]0

2DTFS,Nx n X kπ

Ω =←→

[ ] [ ]02DTFS,Nh n H kπ

Ω =←→

Periodic Convolution (FS)

( ) ( )

[ ] [ ] [ ]T

y(t) h(t) x(t) h x t d

Y k T H k X k

= ∗ = τ − τ τ

= ⋅ ⋅

[ ]02FS,Ty(t) Y kπ

ω =←→[ ]0

2FS,Tx(t) X kπ

ω =←→

[ ]02FS,Th(t) H kπ

ω =←→

Non-periodic Modulation (FT)

( ) ( )

( )( ) ( )

( ) ( ) ( )

j ' t j t

j ' t j t j( ' ) t2 2

j t j t

1 1y( t) z( t) x( t) Z j ' e d ' X j e d2 2

1 1Z j ' e X j e d ' d Z j ' X j e d ' d2 2

1 Z j ' X j( ' e d ' d2

1e Z j ' X j( ' d ' d e Z j X2

∞ ∞ω ω

−∞ −∞

∞ ∞ ∞ ∞ω ω ω +ω

−∞ −∞ −∞ −∞

∞ ∞λ

−∞ −∞

∞ ∞λ λ

−∞ −∞

= ⋅ = ω ω ⋅ ω ωπ π

= ω ω ω ω = ω ω ω ωπ π

= ω λ − ω ω λπ

= ω λ − ω ω λ = λ ∗ π

∫ ∫

∫ ∫ ∫ ∫

∫ ∫

∫ ∫ ( )j d∞

−∞

λ λ∫

( )FTy(t) x(t) z(t) Y j= ⋅ ←→ ω

( )FTz(t) Z j←→ ω

( )FTx(t) X j←→ ω⊗

( ) ( ) ( )FT 1y(t) z( t) x(t) Y j Z j X j2

= ⋅ ←→ ω = ω ∗ ωπ

Non-periodic Modulation (DTFT)

[ ] [ ] [ ]( )DTFT j

y n x n z n

Y e Ω

←→[ ] ( )DTFT jz n Z e Ω←→

[ ] ( )DTFT jx n X e Ω←→⊗

[ ] [ ] [ ] ( ) ( ) ( )DTFT j j j1y n z n x n Y e Z e X e2

Ω Ω Ω= ⋅ ←→ = ∗π

Parceval Relationships( ) [ ]FTx t X j←→ ω

Energy relations for nonperiodic signals:

[ ] ( )DTFT jx n X e Ω←→

( ) [ ]02FS,Tx t X kπ

ω =←→

[ ] [ ]02DTFS,Nx n X kπ

Ω =←→

( ) ( ) ( ) [ ] ( )

[ ] [ ] [ ]

2 j tx

1Pr oof : E x t dt x t x t dt X j e d x t dt2

1 1X j X j d X j d2 2

∞ ∞ ∞ ∞∗ ∗ − ω

−∞ −∞ −∞ −∞

∞ ∞∗

−∞ −∞

= = = ω ω π

= ω ω ω= ω ωπ π

∫ ∫ ∫ ∫

∫ ∫

( ) [ ]2 2x

1E x t dt X j d2

∞ ∞

−∞ −∞

= = ω ωπ∫ ∫

[ ] [ ]2 2x

1P x n X kN

= =∑ ∑

( ) [ ]2 2x

1P x t dt X kT

−∞

= = ∑∫

[ ] ( ) 22 jx

1E x n X e d2

−∞ π

= = Ωπ∑ ∫

Power relations for periodic signals:

Duality

( ) [ ] ( ) [ ]FT FTx t X j X jt 2 x←→ ω ⇔ ←→ π⋅ −ω

[ ] ( ) ( ) [ ]DTFT FS, 1j jtx n X e X e x kΩ←→ ⇔ ←→ −

[ ] [ ] [ ] [ ]0 02 2DTFS, DTFS,N N 1x n X k X n x k

π πΩ = Ω =

←→ ⇔ ←→ −

Time – Bandwidth Product

( ) ( )2X j sin Tω = ωω

( ) ( )1 2 T 2Z j sin sin 2T½ /½ ½

ω ω = = ω ω ω

-3 -2 -1 0 1 2 30

Time t-T T

- 3 - 2 - 1 0 1 2 30

Time t-2T 2T

z(t)=x(½t)

x Xt 2T 2 4Tπ

∆ ⋅∆ω = ⋅ = π z Zt 4T 2 42Tπ

∆ ⋅∆ω = ⋅ = π

t x t dtT

x t dt

−∞∞

−∞

−∞∞

−∞

ω ω ω

∫ w d1B T2

⋅ ≥

Example

General rule

-20 -15 -10 -5 0 5 10 15 20-1

Correlation Functions( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

2FT *ff

FT *fg

f t F j

R f t f t dt F j F j F j

f t F j g t G j

R f t g t dt F j G j

−∞

←→ ω

τ = + τ ←→ ω ω = ω

←→ ω ←→ ω

τ = + τ ←→ ω ω

Auto correlation :

Cross correlation :

Continuous signals:

[ ] ( )[ ] [ ] [ ] ( ) ( ) ( )

[ ] ( ) [ ] ( )[ ] [ ] [ ] ( ) ( )

DTFT j

2DTFT * j j jff

DTFT DTFTj j

DTFT * j jfg

f n F e

R m f n f n m F e F e F e

f n F e g n G e

R m f n g n m F e G e

∞Ω Ω Ω

=−∞

∞Ω Ω

=−∞

←→

= + ←→ =

←→ ←→

= + ←→

Auto correlation :

Cross correlation :

Discrete signals:

Signal ProcessingPart 3.8 Technical Applications of Fourier representations

Magnus Danielsen

Technical Applications of Fourier representations

• Radar spectrum• Filters• Accelerometers• Geophones• Gauss Pulse• AM radio

Radar MeasurementsSpectrum of RF Pulse Train

0 2T0 T T+2T0 t

[ ] ( )0 0jkT 0 0sin kT

P k ek

− ω ω=

[ ] [ ] [ ]1 1S k k 500 k 5002j 2j

=− δ + + δ −

[ ] [ ] [ ] [ ] [ ] ( )

( ) ( )

0 0 0 0

jkT 0 0

j( k 500)T j( k 500)T0 0 0 0

sin kT1 1X k S k P k k 500 k 500 e2 j 2 j k

sin (k 500)T sin (k 500)T1 1e e2 j (k 500) 2 j (k 500)

− ω

− + ω − − ω

ω = ∗ = − δ + + δ − ∗ π

+ ω − ω= − +

+ π − π

[ ] [ ] [ ] [ ] [ ]0 0 0Hint: k k Z k k ' k Z k k ' Z k kδ − ∗ = δ − − = −∑

FiltersLP=Low-pass filter

HP=High-pass filter

BP=Band-pass filter

BS=Band-stopp filter

ω-W W

H(jω)

ω-W W

H(jω)

ωW1 W2

H(jω)

-W2 -W1

ωW1 W2

H(jω)

-W2 W1

H(ejΩ)

H(ejΩ)Ω

-2π - π -W W π 2π

Ω-2π - π -W W π 2π

H(ejΩ)Ω

-2π - π W1 W2 π 2π

H(ejΩ)Ω

-2π - π W1 W2 π 2π

Analogue filter Digital filter

The Accelerometer Equation and the Geophone Equation

0u (t)

22nn 02

d y(t) dy(t) y(t) u (t)dt Q dt

ω+ + ω =

d 1y(t) v(t)dtv(t) voltage ingeophonecoil

= ⋅α

0 0du (t) u (t)dt

2 22nn 02 2

d v(t) dv(t) dv(t) u (t)dt Q dt dt

ω+ +ω = α

( ) ( )( )

( )( )

G2 2n0 0

V j Y j ( j ) ( j )H jU j U j /( j ) ( j ) j

ω ω ⋅ ω α ωω = = =

ωω ω ω ω + ω + ω

( ) ( )( )A

2 2n0n

Y j 1H jU j ( j ) j

ωω = =

ωω ω + ω+ ω

Accelerometer:

Geophone:

Accelerometer Characteristics

d y(t) dy(t) y(t) u(t)dt Q dt

ω+ + ω =

( ) ( )( )A

Y jH j

( j ) jQ

ωω =

=ωω + ω + ω

u(t) = acceleration to be measured y(t) = displacement

Geophone

A. Barzilai, T Van Zandt, T. Pike, S. Manion, T. Kenny, Stanford University and Jet Propulsion Laboratory, AGU 1998)

PermanentMagnet

Spoli av kopartráði

Geofon húsiFjøður

Geophone Characteristics

2 22n 0n2 2

d v(t) dv(t) d u (t)v(t)dt Q dt dt

ω+ +ω =

( ) ( )( )G

V jH j

( j ) jQ

ωω =

α ω=

ωω + ω + ω

Quality factor: Q 1Damping constant: ξ =

0.330u (t) = velocity to be measured

v(t) = measured coil voltage

Gauss Pulse

( ) ( )

d tg(t) e t g t j G jdt 2

djt g t G jd

−= − = − ⋅ ←→ ω⋅ ω

− ⋅ ←→ ωω

FTg(t) G( j )←→ ω

( ) ( )2

dDifferential equation : G j G jd

G( ) c eω

ω = −ω⋅ ωω

ω = ⋅

2 21g(t) dt 1 G( j ) d c 12

∞ ∞

−∞ −∞

= = ω ω → =π∫ ∫

-5 0 50

21g(t) e2

−σ=

1/ et 2 2∆ = σ

-5 0 50

2G( ) eσ ω

−ω = σ

∆ω =σ

FT2 21GeneralGauss pulse : g(t) e G( ) e2

σ ω− −σ= ←→ ω = σ

1/ e 1/ e2Uncertainty relation for Gauss pulse: t 2 2 8∆ ⋅∆ω = σ⋅ =σ

Amplitude Modulated Radio System – AM

( ) ( ) ( ) ( ) ( )

( ) ( )( ) ( )( )

c cj t j tc

1Transmitted signal : r t m t cos t m t e e2

1 1R j M j M j2 2

ω − ω= ω = +

ω = ω − ω + ω + ω

( ) ( ) ( )

( ) ( )( ) ( )( )

( ) ( ) ( )

Received signal: q t r t cos t

1 1Q j R j R j2 21 1 1M j( 2 ) M j M j( 2 )4 2 4

= ω ←→

ω = ω−ω + ω+ω

= ω− ω + ω + ω+ ω

Source Transmitter Channel Receiver

Usually applied AM – mode : m(t) = 1 + m⋅x(t) where x(t) is the source signal

cos(ωct)

r(t)=m(t)·cos(ωct)⊗cos(ωct)

q(t)⊗ LP-filter

y(t)x(t)Source coding

AM Radio - Spectrum

( ) ( )( ) ( ) ( )

( ) ( )( ) ( )( )

( )( ) ( )( )( ) ( )( ) ( )( )( )

c c c c

for positive frequencies for negativ frequencies

Usually applied AM mode: m t 1 m x t 0 m 1

M j j mX j1 1Transmitted signalspectrum : R j M j M j2 2

1 1j mX j j mX j2 2

− = + ⋅ ≤ ≤

ω = δ ω + ω

ω = ω − ω + ω + ω

= δ ω − ω + ω − ω + δ ω + ω + ω + ω

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

x t X jω

Amplitude modulation : m t 1 m x t M j j +m X jω

Double-sideband modulation : m t x t M j X jω

Single-sideband m

←→

− = + ⋅ ←→ ω = δ ω ⋅

− = ←→ ω =

Source signal x t and spectrum X jω :

Source - coded signal m t spectrum M jω :

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )

FTˆodulation : m t x t jx t M j X jω sgn j X jωˆwhere x t is the Hilbert transform of x t .

AM, DSB, and SSB are all amplitude modulations, but frequently only AM is named amplidtude modulation. SSB requires mor

− = + ←→ ω = + ωSSB

e sophisticated modulation method than sketched here.

AM Radio Signals and Spectra

0 0.01 0.02 0.03 0.04 0.05-1

0 0.01 0.02 0.03 0.04 0.05-2

0 0.01 0.02 0.03 0.04 0.05-1

0 0.01 0.02 0.03 0.04 0.05-2

0 0.01 0.02 0.03 0.04 0.05-1

0 0.01 0.02 0.03 0.04 0.05-2

0 500 1000 1500 20000

0.50 500 1000 1500 2000

0 500 1000 1500 20000

0.50 500 1000 1500 2000

0 500 1000 1500 20000

( )R jf

( )X jf

Sine signal Square signal Sawtooth signal

Spectrum (positive frequencies,and m=½ )

Time signal (m=½)

Signal ProcessingPart 4.1 Sampling of Signals

Magnus Danielsen

Sampling and Reconstruction of Analogue Signals

• Sampling • Relations between FT, FS, DTFT & DTFS• Impulse sampling method • Spectral properties of sampled signal• Aliasing • Examples of aliasing• Reconstruction of continuous-time signals• Applications

Sampling

( ) [ ]Sampling at times t nT: x nT x n= =

( )x t

( )( ) ( )FT

Frequency property of x t :

x t X j←→ ω

[ ] [ ] ( )DTFT jFrequency property of x n : x n X e Ω←→

Practical sampling requires that:

• x(t) is described by a number of discrete values x[n] at discrete times t=nTs

• Number of samples N is finite

• x[n] can be periodised, and described by DTFS resulting in a discrete spectrum X[k]

• x(t) has a spectrum described by X(jω)

• The spectrum of x(t) is obtained by 4 relations between FT, FS, DTFT, and DTFS

FT to FS Relation( ) [ ]

( )( )0 0

jk t FT0 0

x t X k

e 2 kω

←→

←→ πδ ω

←→ πδ ω − ω

( ) [ ] ( ) [ ] ( )0jk t FT0

x t X k e X j 2 X k k∞ ∞

=−∞ =−∞

= ←→ ω = π δ ω − ω∑ ∑

0 0jk tFS for e ω

0 0jk tFT for e ω

[ ] [ ]( )

jk n DTFT0

x n X k

1 2 for

e 2 2 mk∞

=−∞

←→

←→ πδ Ω − π < Ω < π

←→ π δ Ω πΩ −−∑

[ ] [ ] ( ) [ ] ( )0

N 1jk n DTFT j

0k 0 k m

x n X k e X e 2 X k 2k m− ∞ ∞

= =−∞ =−∞

−= ←→ = π Ω πδ Ω−∑ ∑ ∑

0 0jk nDTFS for e Ω

0k 0N k+

0N k− +

... ...Ω

0 0jk nDTFT for e Ω

0 0k Ω 0 02 kπ + Ω

0 02 k− π + Ω

... ...2− π

DTFT to DTFS Relation

FT to DTFT Relation[ ] ( )( )

DTFT j

j T nFTs

x n X e

t nT e

←→

δ ←→

δ − ←→

[ ] ( )

( ) [ ] s

FTj T nj

s Tn n

x t X j

x n t nT X e x n e

∞ ∞− ωΩ

Ω=ω=−∞ =−∞

= ω = ←→

δ − = ∑ ∑

T2Sampling interval: T

Ω = ωπ

( )jDTFT : X e Ω

2 π... ...

2− π

( )FT : X jδ ω

... ...

( )x tδ

... ...

[ ]x n

... ...n

[ ] [ ]DTFSx n X k←→

[ ] [ ] ( ) [ ] ( )0

N 1jk n DTFT j

0k 0 k

x n X k e X e 2 X k k− ∞

= =−∞

= ←→ = π δ Ω − Ω∑ ∑

T2Sampling interval: T

Ω = ωπ

( ) ( ) ( )

FT j j

x t X e X e

2 X k T ( k )T

Ω Ωδ δ Ω=ω

=−∞

←→ =

Ω= π δ ω−

[ ]x n

... ...

n( )x tδ

sNTsNT−t

[ ]X k

... ...

( )X jδ ω

... ...

FT to DTFS Relation

Sampling – Impulse Sampling Method

( ) [ ]s sSampled signal at times t nT : x nT x n= =

( )x t

( )( ) ( )FT

x t X j←→ ω

[ ] [ ] ( )DTFT jFrequency property of x n : x n X e Ω←→

( ) ( ) ( )1X j X j P j2δ ω = ω ∗ ωπ

( ) ( )

( ) [ ] ( ) ( ) ( ) ( )

s s sn n

p t t nT

x t x n t nT x(nT ) t nT = x t p t

=−∞

∞ ∞

δ=−∞ =−∞

= δ −

= δ − = δ − ⋅

∑ ∑Impulse sampled signal :

Spectrum of impulsesampled signal :

Spectrum for Sampled Signal (1)

( ) ( ) ( )1X j X j P j2δ ω = ω ∗ ωπ

( )x t

( ) ( ) ( )x t x t p tδ = ⋅

( )p t

......

( ) ( )

[ ] ( )s

p t t nT

1 1P k t e dtT T

=−∞

− ω

= δ − ←→

= δ ⋅ =

( ) ( ) [ ]FTs

n k ks s s

2 2 2p t t nT P( j ) 2 P k k kT T T

∞ ∞ ∞

=−∞ =−∞ =−∞

π π π= δ − ←→ ω = π δ ω − = δ ω −

∑ ∑ ∑

( ) ( ) ( )

( )k ks s s s

1X j X j P j21 2 2 1 2X j k X k

2 T T T T

∞ ∞

=−∞ =−∞

ω = ω ∗ ωπ

π π π= ω ∗ δ ω − = ω − π

∑ ∑

( )x t

( ) ( ) ( )x t x t p tδ =

( )p t

......

( ) ( ) ( )ks s

1 1 2X j X j P j X k2 T T

δ=−∞

πω = ω ∗ ω = ω − π

( )P jω

ω......

ω-W W

X(jω)

ω-W W

Xδ (jω)

ω = s2ωs−ωs2− ω

... ...

Spectrum for Sampled Signal (2)

Sampling with Rectangular Pulses (1)( )x t

( ) ( ) ( )sx t x t p t= ⋅

( )x tτ

( )p t

......τ

( )( ) ( )FT

x t X j←→ ω

( ) [ ] [ ]0jk t FS

p t c k e c k∞

=−∞

= ←→∑

( ) ( ) ( ) [ ] ( )

( ) ( )

[ ] ( )

[ ] ( ) ( )

FT j ts s

jk t j t

x t p t x t c k e x t

X j x t e dt

c k e x t e dt

c k x t e dtc k X( k )

=−∞

∞− ω

−∞

∞ ∞ω − ω

=−∞−∞

∞− ω− ω∞ ∞

−∞=−∞ =−∞

= ⋅ = ⋅

←→ ω =

= ⋅ ⋅

⋅= = ω − ω

∑∫

∫∑ ∑

Spectrum for Rectangular Pulse Sampled Signal

ω-W Ws

ω = s2ωs−ωs2− ω... ...

|Xs(jω)|c[k]=c[ω=kωs]

( ) [ ]s sk

X j c k X( k )∞

=−∞

ω = ω − ω∑

ω-W W

|X(jω)|

The distribution of Xs(jω) on the ω-axis is the same as for the ideally sampled signal

( ) ( )FTx t X j←→ ω

Signal ProcessingPart 4.2 Aliasing and Reconstruction of Sampled Signals

Magnus Danielsen

Aliasing

Nyquist Sampling Theorem:

The condition for reconstruction of a sampled signal x(t) is that the maximum bandwidth W of the sampled signal x(t) shall be less than half of the sampling frequency, or W < ½·ωs = π/Ts.

ωs is called the Nyquist cyclic frequency,

and fs= ωs/2π is the Nyquist sampling frequency.(Geophysisists name fs = ωs/2π the Nyquist frequency)

Xδ (jω)

ω-W Ws

... ...

s3− ω s3ω

Sampling with aliasing error

ω-W W

Xδ (jω)

... ... sW2ω

Sampling with no aliasing error

ω-W W

Xδ (jω)

... ...

s3ωs3− ωsW

Sampling on the limit to aliasing error

DTFT of Sampled Signal Spectrum

[ ] ( )s

DTFT j

A sampled signal having the frequency spectrum X (j ) corresponds to the time signal x (t). Therefore the sampling values are

x n X e X (j )

ΩΩδ ω=

←→ = ω

sWT < πΩ-WTs WTs

2π 4π2− π4− π

... ...

( )jX e Ω

... ...

-WTs WTs2π 4π2− π4− π

( )jX e Ω

sWT = π

sWT > π

... ...

( )jX e Ω

-WTs WTs2π 4π2− π4− π

Example: Sampling a Sinusoidal Signal( ) ( )Signal: x t cos t= π sSampling time: T

- 4 - 3 - 2 - 1 0 1 2 3 4- 1

- 0 . 8

- 0 . 6

- 0 . 4

- 0 . 2

1 ( )x t

- 4 - 3 - 2 - 1 0 1 2 3 4- 1

- 0 . 8

- 0 . 6

- 0 . 4

- 0 . 2

1 ( )x t

• •• • •••

- 4 - 3 - 2 - 1 0 1 2 3 4- 1

- 0 . 8

- 0 . 6

- 0 . 4

- 0 . 2

1 ( )x t

- 4 - 3 - 2 - 1 0 1 2 3 4- 1

- 0 . 8

- 0 . 6

- 0 . 4

- 0 . 2

1 ( )x t

4π 8π8− π 4− π 0 π−π

X ( j )δ ω

4π 8π8− π 4− π 0 π−π

X( j )ω

s 8ω = π

4π 8π8− π 4− π 0 π−π

X ( j )δ ω

..... .....s

4s 3ω = π

4π 8π8− π 4− π 0 π−π

X ( j )δ ω

ωsT 1=

s 2ω = π

..... .....

s 8ω = π

Aliasing of Moving Wheel in Movies

(rad / s)ω

sTϕ = ω< π

sSampling time: T

(rad / s)ω

sTϕ = ω> π

No aliasing error:

Correct apparent rotation direction and angle frequency ωa = ω

Aliasing error :

Wrong apparent rotation direction and wrong apparent angle frequency ωa = (2π - ϕ)/Ts = 2π/Ts - ω

Reconstruction of Continuous-time Signal (1)

ω-W W

Xδ (jω)

... ... sW2ω

Sampling with no aliasing error

ω-W W

X(jω)

H(jω)

sW B2ω

( ) ( ) ( ) ( ) [ ] ( ) [ ] ( )

r r s r s

y t h t x t h t x n t nT x n h t nT

x n sinc (t nT )2

∞ ∞

δ−∞ −∞

−∞

= ∗ = ∗ δ − = −

ω = − π

∑ ∑

( ) ( )FTy t Y j←→ ω( ) ( )FTx t X jδ δ←→ ω ( ) ( )FTr rh t H j←→ ω

Filter with bandwidth B = ωs/2

( ) s s sr

Th t sin t sinc tt 2 2

ω ω = = π π ( )

T for2H j

0 for2

ω ω <ω = ω ω >

-4 -3 -2 -1 0 1 2 3-2

Practical ReconstructionZero-order Hold

Zero-order hold

[ ]y n [ ]0y n

Anti-imaging filter

( )y t

[ ]y n

[ ]0y n( )x t

( )y t

Multipath Communication Channel- Reflections from Environment (1)

[ ] [ ] [ ]( ) ( ) ( )

y n x n a x n 1

Y e X e 1 a eΩ Ω − Ω

= + ⋅ − ←→

= + ⋅

[ ] ( )DTFT jx n X e Ω←→[ ] ( )FT jh n H e Ω←→

Discrete-time model

( ) ( ) ( )( ) ( ) ( )diff

FTdiff

y t x t x t T

Y j X j 1 e− ω

= + α ⋅ − ←→

ω = ω + α

( ) ( )FTx t X j←→ ω( ) ( )FTh t H j←→ ω

Two-path communication channel

( ) ( )( ) ( )diffj TY j

H j 1 eX j

− ωωω = = + α

( ) ( )( ) ( )

Y eH e 1 a e

ΩΩ − Ω

Ω= = + ⋅

( ) ( ) ( )( )

/ T2 2 2 2s

TMSE H j H j d a 12

MSE 1 for a

δ−π

= ω − ω ω = − αγ + α − γπ

= α − γ = αγ

( ) ( ) ( )s

TH j H e 1 a e− ωΩ

δ Ω=ωω = = + ⋅⇒

( ) ( )

diff s

/ T2j T j Ts

TMSE H j H j d2

T e a e d2

a a * * a *

δ−π

π− ω − ω

= ω − ω ωπ

= α − ⋅ ωπ

= α + − α γ − α γ

= − αγ + α − γ

( ) ( )sj TH j 1 a e− ωδ ω = + ⋅( ) ( )diffj TH j 1 e− ωω = + α

( )2 2

minMSE 1 for a= α − γ = αγ

diff s

/Tj (T T )s diff s

T T Te d sinc2 T

π− ω −

−γ = ω= π

0.5 1 1.50

21− γ

Tdiff/Ts

Multipath Communication Channel− Reflections from Environment (2)

Basic Discrete-time Signal Processing System for Analogue Signals

Anti-aliasing filter

Sampling at interv.Ts

Discrete-time processing

Sample and hold

Anti-imaging filter

( )x t ( )ax t [ ]x n [ ]y n ( )0y t ( )y t

Equivalent continuous-time system G(jω)

( )x t ( )y t

Signal ProcessingPart 5.1 Discrete Fourier Transform (DTF)

for nNumerical Evaluation

Magnus Danielsen

Discret Forier Transform (DTF) and Fast Fourier Transform (FFT)

• DTFS is the only Fourier representation capable of exactnumerical evaluation

• DTFT related to DTFS• FS related to DTFS• FT related to DTFS• DFT = Discrete Fourier Transform – definition and reasons• FFT = Fast Fourier Transform

= efficient algorithms for calculating DTFS• FFT with decimation in time• FFT with decimation in frequency• FFT and IFFT use nearly same algoritme

Periodicing of Finite Signal Duration

[ ]x n

2Nperiod

[ ] [ ]

[ ] ( ) [ ]

M 1DTFT j j n

n 0N 1

Finite duration signal :

x n for 0 n M 1x n

0 for n 0 n M M N

x n X e x n e

−Ω − Ω

−− Ω

≤ < −=

< ∨ ≥ ≤

←→ =

[ ]x n

[ ] [ ] [ ]

[ ] [ ] [ ] 0

periodisedm

N 1jk nDTFS

Periodiced signal :

x n x n x n mN

1x n X k x n eN

=−∞

−− Ω

←→ =

DTFT Related to DTFS

[ ] [ ] [ ]

[ ] [ ] [ ] 0

periodisedm

N 1jk nDTFS

Periodiced signal :

x n x n x n mN

1x n X k x n eN

=−∞

−− Ω

←→ =

[ ] [ ]

[ ] ( ) ( ) [ ]N 1

DTFTj j n j j n

x n for 0 n N 1x n

0 for n 0 n N

π −Ω Ω Ω − Ω

=−π

≤ < −=

< ∨ ≥

= Ω←→ =π ∑∫

[ ] ( ) ( )0

1 1X k X e X eN N

ΩΩΩ= Ω= =

[ ] ( ) ( ) [ ]0 0 0

N 1 N 1j k j kn j knj j n

k 0 k 0

1 1x n X e e d X e e X k e2 N

π − −Ω − Ω − ΩΩ Ω

= =− π

= Ω ≅ =π ∑ ∑∫

Some relations :2dN

πΩ = Ω =

Ω = Ω

[ ] ( ) [ ] [ ] 0

N 1jk nDTFS

disc s discn 0

Discrete periodic signal (sampled signal) :

1x n x nT X k x n eN

−− Ω

= ←→ = ∑

( ) ( ) [ ] ( ) [ ]0

T Njk t jk nTFS

0 sn 00 00

Periodic continuous signal :

1 1x t x t T X k x t e dt x n e TT T

− ω − ω

= + ←→ = ≅ ∑∫

[ ] [ ]0 s 0

2disc TN

X k X k πω =Ω =

0 s 0 0 s s s 0 0 s 02Some relations : T NT T 2 T 2 N TNπ

= ω = π ω = π ω = ω ω = Ω =

FS Related to DTFS

[ ] ( ) [ ] [ ] 0

N 1jk nDTFS

disc s discn 0

Discrete periodic signal (sampled signal) :

1x n x nT X k x n eN

−− Ω

= ←→ = ∑

( ) ( )

( ) ( ) ( )0

TFT j t

x t for 0 t Tx t

0 for t 0 t T

x t X j x t e dt− ω

≤ <= < ∨ ≥

←→ ω = ∫

[ ] [ ] [ ]0 s 0

s2disc TN 0 s

1 1X k X k X k X jkT NT N

πω =Ω =

ω = = =

( ) ( ) ( )

( ) [ ] ( )0

0periodisedmT

jk tFS

Periodiced signal :

x t x t x t mT

1x t X k x t e dtT

=−∞

− ω

←→ =

0 s 0 0 s s s 0 0 s 02Some relations : T NT T 2 T 2 N TNπ

= ω = π ω = π ω = ω ω = Ω =

FT Related to DTFS

Discrete Fourier Transform (DFT) –Definition and Reasons

• DTFS is the only Fourier representation capable of exactnumerical evaluation

• DTFS (period N) relates directly to DTFT (finite time interval N)

• DTFT represents (nonperiodic) energy signals• DTFS represents (periodic) power signals• Nonperiodic signals restrictet to a time interval N, and

correspondingly periodiced signals with period N, energy E and power P are proportional (E=P·N). Correspondingly ”XDTFT=N · XDTFS”

• Resonably Discrete Fourier Transform (DFT) for a finite interval (N) signal is defined as a transform of an energy signal as N times the DTFS

• Consequently DFT is a numerical discrete method to calculate the DTFT for an energy signal.

DFT Formulation for Calculation of DTFT

[ ] [ ] [ ] [ ]0 0

N 1 N 1jk n jk nDTFS

k 0 n 0

Periodiced signal, and DTFS :

− −− Ω − Ω

= ←→ =∑ ∑

[ ] ( ) ( ) [ ]N 1

DTFTj j n j j n

Finite duration signal, and DTFT :

π −Ω Ω Ω − Ω

=−π

= Ω←→ =π ∑∫

[ ] [ ] [ ] [ ]0 0

N 1 N 1jk n jk nDFT

k 0 n 0

Discrete Fourier Transform(DFT)formulation

for calculation of DTFT of finite duration signal :

− −Ω − Ω

= ←→ =∑ ∑

Matrix Formulation of DFT (1)[ ] [ ] [ ] [ ]0 0

N 1 N 1jk n jk nDFT

k 0 n 0

− −Ω − Ω

= ←→ =∑ ∑

[ ][ ][ ][ ]

0 0 0 0 0

j j2 j3 jk j(N 1)

j2 j4 j6 j2k j2(N 1)

j3 j6 j9 n j3k j3(N 1)

1 1 1 1 ... 1 ... 1x 01 e e e ... e ... ex 11 e e e ... e ... ex 21 e e e ... e ... ex 3 1... ... ... ... ... ... ... ...... N1 ex n

...x N 1

Ω Ω Ω Ω − Ω

= ⋅ −

[ ][ ][ ][ ]

0 0 0 0 0

20 0 0 0 0

jn jn2 jn3 jnk jn(N 1)

j(N 1) j(N 1)2 j(N 1)3 j(N 1)k j(N 1)

e e ... e ... e... ... ... ... ... ... ... ...

1 e e e ... e ... e

Ω Ω Ω Ω − Ω

− Ω − Ω − Ω − Ω − Ω

X 0X 1X 2X 3...X k...

X N - 1

[ ][ ][ ][ ]

0 0 0 0 0

j j2 j3 jk j(N 1)

j2 j4 j6 j2k j2(N 1)

j3 j6 j9 n j3k j3(N 1)

1 1 1 1 ... 1 ... 11 e e e ... e ... e1 e e e ... e ... e1 e e e ... e ... e... ... ... ... ..

− Ω − Ω − Ω − Ω − − Ω

X 0X 1X 2X 3...X k...

X N - 1

[ ][ ][ ][ ]

0 0 0 0 0

20 0 0 0 0

jk j2k j3k jnk j(N 1)k

j(N 1) j2(N 1) j3(N 1) jn(N 1) j(N 1)

x 0x 1x 2x 3

. ... ... ... ...1 e e e ... e ... e x n... ... ... ... ... ... ... ... ...

x N 11 e e e ... e ... e

− Ω − Ω − Ω − Ω − − Ω

− − Ω − − Ω − − Ω − − Ω − − Ω

[ ] [ ]DFT

x n X k

X W x1x W XN

←→

= ⋅∗

= ⋅ ⋅

0 0 0 0 0

j j2 j3 jk j(N 1)

j2 j4 j6 j2k j2(N 1)

j3 j6 j9 n j3k j3(N 1)

jn jn2 jn3 jnk jn(N 1)

1 1 1 1 ... 1 ... 11 e e e ... e ... e1 e e e ... e ... e1 e e e ... e ... e

W ... ... ... ... ... ... ... ...1 e e e ... e ... e... ... ..

Ω Ω Ω Ω − Ω

20 0 0 0 0j(N 1) j(N 1)2 j(N 1)3 j(N 1)k j(N 1)

. ... ... ... ... ...

1 e e e ... e ... e− Ω − Ω − Ω − Ω − Ω

0 0 0 0 0

j j2 j3 jk j(N 1)

j2 j4 j6 j2k j2(N 1)

j3 j6 j9 n j3k j3(N 1)

jk j2k j3k jnk

1 1 1 1 ... 1 ... 11 e e e ... e ... e1 e e e ... e ... e1 e e e ... e ... e

W ... ... ... ... ... ... ... ...1 e e e ... e

− Ω − Ω − Ω − Ω − − Ω

− Ω − Ω − Ω − Ω

20 0 0 0 0

j(N 1)k

j(N 1) j2(N 1) j3(N 1) jn(N 1) j(N 1)

... e... ... ... ... ... ... ... ...

1 e e e ... e ... e

− − Ω

− − Ω − − Ω − − Ω − − Ω − − Ω

N N N1 W W UNunity matrix

∗ ⋅ ⋅ =

Matrix Formulation of DFT (2)

Matrix Formulation of DFT (3)[ ] [ ] [ ] [ ] [ ] [ ]0 0

N 1 N 1 N 1 N 1jk n jk nDFTkn kn

k 0 k 0 n 0 n 0

1 1x n X k e X k w X k x n e x n wN N

− − − −Ω − Ω−

= = = =

= = ←→ = =∑ ∑ ∑ ∑

[ ][ ][ ][ ]

( )N 11 2 3 k

2 4 6 2k 2(N 1)

3 6 9 3k 3(N 1)

n n 2 n 3 n k n(N 1)

1 1 1 1 ... 1 ... 1x 0x 1 1 w w w ... w ... wx 2 1 w w w ... w ... wx 3 1 w w w ... w ... w1... ... ... ... ... ... ... ... ...Nx n 1 w w w ... w ... w

....x N 1

− −− − − −

− − − − − −

− − ⋅ − ⋅ − ⋅ − −

= ⋅ −

[ ][ ][ ][ ]

[ ]2N 1 (N 1) 2 (N 1) 3 (N 1) k (N 1)

.. ... ... ... ... ... ... ...

1 w w w ... w ... w− − − − ⋅ − − ⋅ − − ⋅ − −

X 0X 1X 2X 3...X k...

X N -1

[ ][ ][ ][ ]

1 2 3 n (N 1)

2 4 6 2n 2(N 1)

3 6 9 3n 3(N 1)

k k 2 k 3 k n k(N 1)

1 1 1 1 ... 1 ... 11 w w w ... w ... w1 w w w ... w ... w1 w w w ... w ... w... ... ... ... ... ... ... ...1 w w w ... w ... w... ... ... ... ... ... ... .

⋅ ⋅ ⋅ −

X 0X 1X 2X 3...X k...

X N - 1

[ ][ ][ ][ ]

[ ]2(N 1) (N 1) 2 (N 1) 3 (N 1) n (N 1)

x 0x 1x 2x 3...x n

.. ...x N 11 w w w ... w ... w− − ⋅ − ⋅ − ⋅ −

− Ω

[ ] [ ]DFT

x n X k

X W x1x W XN

←→

= ⋅∗

N 11 2 3 k

2 4 6 2k 2(N 1)

3 6 9 3k 3(N 1)

n n 2 n 3 n k n(N 1)

N 1 (N 1) 2 (N 1)

1 1 1 1 ... 1 ... 1

1 w w w ... w ... w1 w w w ... w ... w1 w w w ... w ... w

W... ... ... ... ... ... ... ...1 w w w ... w ... w... ... ... ... ... ... ... ...

1 w w w

− −− − − −

− − − − − −

− − ⋅ − ⋅ − ⋅ − −

− − − − ⋅ − − ⋅

23 (N 1) k (N 1)... w ... w− − ⋅ − −

1 2 3 k (N 1)

2 4 6 2k 2(N 1)

3 6 9 3k 3(N 1)

n n 2 n 3 n k n(N 1)

(N 1) (N 1) 2 (N 1) 3 (N 1) k (N 1

1 1 1 1 ... 1 ... 11 w w w ... w ... w1 w w w ... w ... w1 w w w ... w ... w

W ... ... ... ... ... ... ... ...1 w w w ... w ... w... ... ... ... ... ... ... ...

1 w w w ... w ... w

⋅ ⋅ ⋅ −

− − ⋅ − ⋅ − ⋅ −

1 W WNU unity matrix

− ⋅ ⋅

Matrix Formulation of DFT (4)

Signal ProcessingPart 5.2 Fast Fourier Transform (FFT) – an Efficient

Algorithm for Numerical Evaluation of DFT

Magnus Danielsen

Number of Addition and Multiplications

[ ] [ ] [ ] [ ]0 0

N 1 N 1jk n jk nDFT

k 0 n 0

− −Ω − Ω

= ←→ =∑ ∑

Using DTFS directly•Number of complex multiplication = (N-1)2

•Number of complex additions = N(N – 1)

Using FFT algorithms :•Number of complex multiplication = (N/2) log2(N/2)•Number of complex additions = N log2(N)

2jj NNw w e e

π−− Ω= = =

[ ][ ][ ][ ]

[ ][ ]

1 2 3 (N 1)

2 4 6 2(N 1)

3(N 1)3 6 9

N / 2 1 N / 2 1 2 N / 2 1 3 N / 2 1 (N 1)

N / 2 N / 2 2 N / 2 3 N / 2 (N 1)

1 1 1 1 ... 11 w w w ... w1 w w w ... w1 w w w ... w... ... ... ... ... ...

1 w w w w...

1 w w w w... .

− Ω

− − − − −

X 0X 1X 2X 3...

X N/2 - 1

X N/2...

X N - 1

[ ][ ][ ][ ]

[ ][ ]

[ ]2(N 1) (N 1)2 (N 1)3 (N 1)

.. ... ... ... ... ...

1 w w w ... w

x N/2 -1

-1− − − −

N / 2 d N / 2

⋅ = ⋅

oddind

evenindN ex

N NNw w 1= =

1N / 2

2N / 2

3dN / 2

N / 2 1N / 2

1 0 0 0 ... 00 w 0 0 ... 00 0 w 0 ... 0

W 0 0 0 w ... 0... ... ... ... ... ...

0 0 0 0 ... w −

pnpnW w=

FFT Algorithm – Decimation in Time

2j 2 2N / 2N / 2 Nw e w w

FFT Algorithme – Decimation in TimeFourier Transform X[k] for N=8

⊗1N / 2w

1N / 2w

⊗0Nw

⊗1Nw

⊗2Nw

⊗3Nw

0N / 2w

2jj NNw w e e

π−− Ω= = =

N/2 N/2 d

⋅ = ⋅

evenindex

ind xXX

N NNw w 1= =

3dN/ 2

N/ 2 1N/ 2

1 0 0 0 ... 00 w 0 0 ... 00 0 w 0 ... 0

W 0 0 0 w ... 0... ... ... ... ... ...

0 0 0 0 ... w −

pnpnW w=

FFT Algorithm – Decimation in Frequency

2j 2 2N / 2N / 2 Nw e w w

[ ][ ][ ][ ]

[ ][ ]

1 (N/2 1) N/2 (N 1)

2 2 (N/2 1) 2 N/2 2(N 1)

3 3(N/2 1) 3 N/2 3(N 1)

N/2 1 N/2 1 (N/2 1)

1 1 1 1 ... 11 w w w ... w1 w w w ... w1 w w w ... w... ... .... ... ... ... ...

1 w w1 w

− −

⋅ − ⋅ −

− − ⋅ −

X ½N-1...

..X N-1

…( )

[ ][ ][ ][ ]

[ ][ ]

N/2 1 N/2 N/2 1 (N 1)

N/2 (N/2 1) N/2 (N/2) N/2 (N 1)

(N 1) (N 1) (N/2 1) (N 1) N/2 (N 1)

x 0x 1x 2

w w x ½N -1...

x ½Nw w w... ... ... ... ... ... ...

x N -11 w w w ... w

− ⋅ − −

⋅ − ⋅ −

− − ⋅ − − ⋅ −

FFT Algorithme – Decimation in FrequencyFourier Transform X[k] for N=8

111 111

⊗0Nw

⊗1Nw⊗

⊗3Nw

−⊗1N / 2w

⊗0N / 2w

⊗1N / 2w

2jj NNw w e e

π−− Ω= = =

[ ][ ][ ][ ]

[ ][ ]

1 2 3 (N 1)

2 4 6 2(N 1)

j3(N 1)3 6 9

N / 2 1 N / 2 1 2 N / 2 1 3

N / 2 N / 2 2 N / 2 3

1 1 1 1 ... 1x 01 w w w ... wx 11 w w w ... wx 21 w w w ... wx 3

1 ... ... ... ... ... ......N 1 w w w wx N / 2 1

...1x N / 2 w w w

...x N 1

− − − − −

− − Ω− − −

− − − − − − −

− − −

= ⋅ − −

[ ][ ][ ][ ]

[ ][ ]

N / 2 1 (N 1)

N / 2 (N 1)

(N 1) (N 1)2 (N 1)3 (N 1)

w... ... ... ... ... ...

1 w w w ... w

− −

− − − − − − − −

X N/2 - 1

N/2 d N/2

W W W1N

∗ ∗ ∗ ⋅ = ⋅

∗ ∗ ∗ ⋅

evenin

oddindex

N / 2x

N NNw w 1= =

1N / 2

2N / 2

3dN / 2

N / 2 1N / 2

1 0 0 0 ... 00 w 0 0 ... 00 0 w 0 ... 0

W 0 0 0 w ... 0... ... ... ... ... ...

0 0 0 0 ... w

− −

pnpnW w=

IFFT Algorithm – Decimation in Frequency

2j 2 2N / 2N / 2 Nw e w w

IFFT Algorithme – Decimation in FrequencyTime Function x[n] for N=8

N / 2w−

1N / 2w−

1Nw−

2Nw−

3Nw−

0N / 2w

⊗1N⊗

2jj NNw w e e

π−− Ω= = =

oddindex

N/2 N/2 d

evenindex

W W W1N

∗ ∗ ∗ ⋅ = ⋅

∗ ∗ ∗ ⋅

N NNw w 1= =

3dN/ 2

N/2 1N/2

1 0 0 0 ... 00 w 0 0 ... 00 0 w 0 ... 0

W 0 0 0 w ... 0... ... ... ... ... ...

0 0 0 0 ... w

− −

pnpnW w=

IFFT Algorithm – Decimation in Time

2j 2 2N / 2N / 2 Nw e w w

[ ][ ][ ][ ]

[ ][ ]

1 (N/2 1) N/2 (N 1)

2 2 (N/2 1) 2 N/2 2(N 1)

3 3(N/2 1) 3 N/2 3(N 1)

x ½N 1

1 1 1 1 ... 11 w w w ... w1 w w w ... w1 w w w ... w

1 ... ... .... ... ... ... ......N 1 w

− − − − − −

− − ⋅ − − ⋅ − −

− −

[ ][ ][ ][ ]

[ ][ ]

N/2 1 (N/2 1) N/2 1 N/2 N/2 1 (N 1)

N/2 (N/2 1) N/2 (N/2) N/2 (N 1)

(N 1) (N 1) (N/2 1) (N 1) N/2 (N 1)

w w w...

w w w... ... ... ... ... ...

1 w w w ... w

− − ⋅ − − − ⋅ − − −

− ⋅ − − ⋅ − −

− − − − ⋅ − − − ⋅ − −

X 0X 1X 2X 3...X ½N-1

X ½N...

IFFT Algorithme – Decimation in TimeTime Function x[n] for N=8

⊗0Nw

−⊗1

N / 2w−

⊗0N / 2w

N / 2w−

⊗1N⊗

Matlab for Calculation of DFT (FFT)• Matlab Fourier representation is based on

DTF (FFT) calculations• Spectrum calculation of x(t):

– Define sampling interval Ts– Define t-range T for sampling (e.g. t=[0:Ts:T])– Number of samples N=T/Ts– Define corresponding frequency range (e.g.

f=[0:1/T:1/Ts] (or cyclic frequency ω=2πf)– Matlab command X=fft(x) calculates N DFT-samples– Spectrum of x is obtained:

• Magnitude spectrum: plot(f,abs(X))• Phase spectrum: plot(f,angle(X))

• Time function calculation of X(f):– Matlab command x=ifft(X) calculates N samples of

IDTF– x is generally a complex number, but usually very

near real, when x(t) represents a physical quantity – Time function is therefore found by x(t)=real(x)– A measure of approximation error is the imaginary

part denoted by imag(x)

Example of fft calculation:Ts=0.01;T=20;N=T/Ts;t=0:Ts:T;f=0:1/T:1/Ts;w=2*pi*f;x=exp(-t);X=fft(x);subplot(2,1,1);plot(f(1:50),abs(X(1:50)));subplot(2,1,2);plot(f(1:50),angle(X(1:50)));

Useful Matlab Commands

For vectors of sampled values x and y of time dependent physical functions x(t) and y(t) we can calculate:

• Convolution: x∗y=conv(x,y)• Crosscorrelation: Rxy=xcorr(x,y)• Autocorrelation: Rxx=xcorr(x,x)• p-multiple zero padding when FFT: fft(x,p)• p-multiple zero padding when IFFT: ifft(x,p)• Performing DTFS: x[n]=N·ifft(X)↔X[k]=1/N·fft(x)

Signal ProcessingPart 6.1 Laplace Transform Principles

Magnus Danielsen

Laplace Transform ( ) ( ) ( )

( ) ( ) ( )

Bilateral (double sided) : x t X s x t e dt

Unilateral (single sided) x t X s x t e dt

∞−

−∞

∞−

←→ =

( ) ( )0

Inverse Laplace Transform x t X s e dtσ + ∞

σ − ∞

s-plane

jω s=σ+jω

Integration loop for evaluation of inverse Laplace transform

Fourier transform: s = jω ⇒ X(s) = X(jω)

Eigenvalue Property of est

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

s t st s st

Operator formulation : y t H x t

Convolution : y t H x t h t x t h x t d

h e d e h e d e H s

Eigenfunction : x t eEigenvalue : H(s)

Transfer function eigenvalue : H s h e d

−∞

∞ ∞−τ − τ

−∞ −∞

∞− τ

−∞

= = ∗ = τ − τ τ

= τ τ = τ τ =

= = τ τ

∫ ∫

H( ) stx t e= ( ) ( ) ( )sty t H x t e H s= =

Region of Convergence - ROC

( ) ( ) ( )L stt 0

Causal signal : x(t) 0 x t X s x t e dt∞

−<= ⇒ ←→ = ∫

ROC: Re(s)>σ0

jω s-plane

( ) ( ) ( )Lat at st

Example: Causal exponential signal:

1x t e u t e u t e dts a

ROC : Re(s) a

∞−

= ←→ =−

( ) ( ) ( )0

L stt 0

Anti causal signal : x(t) 0 x t X s x t e dt−>

−∞

− = ⇒ ←→ = ∫

ROC: Re(s)<σ0

jω s-plane

( ) ( ) ( )0

Lat at st

Example: Anti-causal exponential signal:

1x t e u t e u t e dts a

ROC : Re(s) a

−∞

= − ←→ − =−

Unilateral Laplace Transforms Relations for Causal Signals

( ) ( ) ( ) ( )

( ) ( )( ) ( )

( ) ( ) ( ) ( )

s t L0

• Linearity ax t by t aX s bY s

1 s• Scaling x at Xa a

• Time shift x t a e X s

• s-domain shift e x t X s s

• Convolution x t y t X s Y s

• Diffe

+ ←→ +

←→

− ←→ ⋅

←→⋅ −

∗ ←→ ⋅

( ) ( )

( ) ( ) ( )

( ) ( )

drentiation in s-domain t x t X sds

d• Differentiation in time x t s X s x 0dt

1• Integration x t dt X ss

• Initial value theorem lim (s X s ) x 0

• Final val

+→∞

− ⋅ ←→

←→ ⋅ −

←→

( ) ( )s 0

ue theorem lim (s X s ) x→

⋅ = ∞

( ) ( )( ) ( )

x t X s

y t Y s

←→

Given Laplace transforms :

Some Basic Transform Pairs and use of Integral Relation

: (t) 1

1: u(t) t dts

1: t u(t) u t dts

1 1t u(t) t u t dt2 s

t 1u(t)n - 1 ! s

δ ←→

= δ ←→

⋅ = ←→

⋅ = ⋅ ←→

⋅ ←→

Delta function

Step function

Ramp function

Inverse Laplace Transforms of a Rational Function

( )( )

k kMM M 1 m

kM M 1 m 0 k 1NN N 1 n

N N 1 n 0 kk 1

in engineering

is of the form of a rational function w ith real coefficients a and b :

s cb s b s ... b s bX (s)a s a s ... a s a s d

−− =

−+ + + += =

+ + + + −∏∏

C om m only occuring Laplace transform

N 1 N 1' k ' kM N M Nk kk kk 0 k 0

k kN Nk 0 k 0kk kk 0 k 1

M N R N Pk k,rk rk 0 r 1 k 1

with real and complex poles into a sum of simpler function:

b s b sX(s) s s

a s s d

es where R is the mu

− −− −= == =

− −

= α + = α +−

= α +−

∑ ∑∑ ∑∑ ∏

∑ ∑ ∑

Decomposition of X(s)

ltiplicity (order) of the poles.

is obtained using

inverse Laplace transform for each of these simpler terms

The inverse Laplace transform x(t) of X(s)

M re a l o r com p lex ze ros c , w h e re th e n om in a tor is z e ro

N rea l o r c om p lex p o le s d , w h e re th e d en om in a to r is z e ro C om p lex p o le s an d ze ros occu r in com p lex p a ir w h en p re sen t

••

X (s) h a s :

Alternative Decomposition With Complex Pol-pairs Present

N 1 ' kM N M Nkk kk 0

k kNk 0 k 0kkk 0

with real and complex poles, into a sum of simpler function

The complex poles are paired giving quadric forms in the denominator.

b sX(s) s s

−− −== =

= α + = α∑∑ ∑∑

Decomposition of X(s)

( ) ( )

N 1 ' kkk 0

2k k k

real complexpoles pole pairs

R RM N k k,r k,r k,r

k r rk 0 2r 1 k 1 r 1 k 1k k kand real and complex

poles only pole pairs only

s d s s

e f s gs

s d s s

where R is the multiplicity (order) of the poles.

== = = =

+− ⋅ + γ + ζ

+= α + +

− + γ + ζ

∑∏ ∏

∑ ∑ ∑ ∑ ∑

Th is obtained using

inverse Laplace transform for each of these simpler terms

e inverse Laplace transform of X(s)

( )( )

( ) ( )( )

( ) ( )( )( )

L-a t22 2

2 2L-a t

2 st sin ts

st cos ts

1 2t cos t sin ts

2 s ae t sin t

s ae t cos t

ω⋅ ω ←→

− ω⋅ ω ←→

− ω⋅ ω − ω ←→

ω + ω

ω +⋅ ⋅ ω ←→

+ + ω

+ − ω⋅ ⋅ ω ←→

+ + ω

2 m ultiple com plex pole pair = ±jω

2 m ultiple com plex pole pair = -a ± jω :

( ) ( )

( )( )

1t cos t sin t

⋅ ⋅ ω − ω ←→ ω − ω

+ + ω

( ) ( )

n 1Ld t

1e u ( t )s d

1t e u ( t )s d

t 1e u ( t )n 1 ! s d

sin ts

scos ts

− ⋅

−− ⋅

⋅ ←→−

⋅ ⋅ ←→−

⋅ ⋅ ←→− −

ωω ←→

ω ←→+ ω

Single pole, real or com plex = d :

2 - m ultiple pole = d :

n - m ultiple pole = d :

C om plex pole pair = ±jω :

C om plex pole pair = -a ± jω :

( )( )

( ) ( )( )

La t2 2

e sin ts a

s ae cos t

− ⋅

ω⋅ ω ←→

+ + ω

+⋅ ω ←→

+ + ω

Laplace Transform Pairs of some Basic Functions

Examples of Decomposition of Laplace Transforms:

( ) ( ) ( ) ( ) ( ) ( )( ) t 2t 3t

3s 5 1 1 2: X(s)s 1 s 2 s 3 s 1 s 2 s 3

x t e e 2 e− − −

+= = + −

+ + + + + +

= + − ⋅

Three single real poles

( ) ( ) ( ) ( )

( )( )( )( ) ( )

( )( )( ) ( )( )

( ) ( ) ( )

t ½t ½t

s 2 1 s 1 X(s)s 1s 1 s s 1 s s 1

s ½ s ½1 1 ¾3s 1 s 1s ½ ¾ s ½ ¾ s ½ ¾

x t e e cos ¾t e sin ¾t− − −

+ −= = −

++ + + + +

+ − += − = − +

+ ++ + + + + +

= − +

1 real pole and 1 complex pole - pair :

( ) ( ) ( ) ( ) ( )( )

2t 4t 4t

3s 10 1 1 1 X(s)s 2 s 4s 2 s 4 s 4

x t e e t e− − −

+= = − +

+ ++ + +

= − + ⋅

Multiple real poles :

Solving Differential Equations with Initial Conditions

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( ) ( )2 t 5t

d 1 1y t y t x tdt RC RCd y t 5y t 5x tdtsY s y 0 5Y s 5X s

3 15 25X(s) y(0 ) 5 s 2Y ss 5 s 5

2s 1 1 3s 5 s 2 s 2 s 5

y t (e 3e )u t

− −

− + =

−+ += =+ +

− += = −

+ + + +

C=200µ

L2t3 3 1x t e5 5 s 2

RC 1k 200µy 0 2

= ←→+

τ = = ⋅

- 1 - 0 . 5 0 0 . 5 1 1 . 5 20

0.6x(t)

- 1 - 0 . 5 0 0 . 5 1 1 . 5 2- 2

- 1 . 5

- 0 . 5

Laplace Transforms in Circuit Analysis - Impedance and Initial Generators

• Resistance

• CapacitanceInitial voltage = VC(0-)

• InductanceInitial current = IL(0-)

+ VR (s) –

IR (s)

+ VC (s) –

IC (s) 1sC

Cv (0 )s

+ VL (s) –

IL (s)

LLI (0 )−

+ VC (s) –

IC (s)

CC v (0 )−⋅

+ VL (s) –

IL (s)

Li (0 )s

+ VR (s) –

IR (s)

Example of Laplace Transformed Circuit(Low-pass Filter)

1Cv (0 ) 0.5s s

2Cv (0 ) 0.3s s

LI (0 ) L 0.1 L− ⋅ = ⋅

+−( )

Bilateral Laplace Transform Properties

• Definition: x(t) for − ∞ < t < ∞• ROC must be given for the transform• x(t) found from X(s) depends of given ROC• Calculation rules same as for unilateral exept

differentiation, where x(0-) does not appear: dx(t)/dt has the transform sX(s)

• Care must be taken to ROC.

Example of Bilateral Laplace Transformand Inverted Time Function

( )( ) ( ) ( ) ( ) ( )3

2 2 2 2

4s 26s 1 1 1 1: X(s)s 2 s 2 s 3 s 3s 2 s 3

( ) : 2 Re(s) 2

−= = + + +

+ − + −− −

− < < +

Laplace transform

Regionof convergence ROC

2t 2t 3t 3tx(t) e u(t) e u( t) e u(t) e u( t)− −= + − + + −

Inverted Laplace transform time function :

First and third term are causal Second and forth term are anti-causal

x x x x

3-2 2-3

x = poless-plane

Signal ProcessingPart 6.2 The Laplace Transform – Transfer Functions

Magnus Danielsen

Transfer Function• Definition of transfer function H(s)=Y(s)/X(s), where Y(s)

and X(s) are output and input signals respectively• Initial conditions are zero• Causality – unilateral transforms• Stability – all poles are in ”negative” half of s-plane• Minimum phase – all zeros are in ”negative” halfplane• Inverse system transfer function H-1(s) of H(s) such that

H-1(s)·H(s)=1• Frequency response determined by poles and zeros• Bode diagrammes• Decibel – definition:

– A=20 logH ,where H is a transfer function. – A=10 log(P2/P1), where P2 and P1 are the output and input powers

respectively.

Transfer Function of System Definition and Differential Equation

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )M M 1 N N 1

M M 1 0 N N 1 1 0M M 1 N N 1

d d d d d db y t b y t ... b y t b y t a x t a x t ... a x t a x tdt dt dt dt dt dt

− −

−− −+ + + + = + + + +

Differential equation :

( )( )

MM M 1 mkM M 1 m 0 k 1

NN N 1 nN N 1 n 0 kk 1

s cY(s) b s b s ... b s bH (s) KX(s) a s a s ... a s a s d

−− =

−+ + + += = = ⋅

+ + + + −∏∏

Definition of transferfunction :

( )M 1 N 1 N 2

M 1 N 1 N 2

d d d d d For it applies that: y ... y y t 0 and x x ... x x 0dt dt dt dt dt

− − −

− − − −= = = = = = = = =

Initial conditions :

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )M M 1 N N 1M M 1 1 0 N N 1 1 0b s Y s b s Y s ... b sY s b Y s a s X s a s X s ... a sX s a X s− −

− −+ + + + = + + + +

Laplace transform of differential equation :

H(s) Lx(t) X(s)←→ Ly(t) Y(s)←→

Transfer Function and Impulse Response

( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

Y s H s X s

y t h t x t , where h t H s

x t t y t h t t h t

= ∗ ←→

= δ ⇒ = ∗δ =

Laplace transform of output signal :

Output signal time function :

Impulse response h(t) :

Lx(t) X(s)←→ Ly(t) Y(s)←→( ) ( )Lh t H s←→

( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )

1Example on impulse response : H ss a

x t t X s 1

Y s H s X s H s

y t h t exp at

= δ ←→ =

= = −

Stability of a SystemFor a transfer function H(s) with impulse response h(t) holds

• Stable systems:– H(s) contains poles in negative half s-plane only– h(t) contains only terms decaying exponentially with time*

– The system is stable• Oscillating unstable systems:

– H(s) contains poles on the imaginary axis of the s-plane– h(t) contains oscillating terms versus time t– The system is oscillating unstable

• Unstable systems:– H(s) contains poles in positive half s-plane – h(t) contains exponentially increasing terms versus time t– The system is unstable

(*impulses can also be present)

Transfer Function -Inverted Transfer Function

( ) ( )Lh t H s←→ ( ) ( )Ly t Y s←→( ) ( )Lx t X s←→

( ) ( )( )

k kk 0 k 1N N

k 0 k 1

b s s cY sH s K

X s a s s d

−= = = ⋅

∑ ∏

( ) ( )k kN M

k kk kk 0 k 0

d da y t b x tdt dt= =

=∑ ∑ ( ) ( ) ( ) ( )k k

L Lk kk k

All initial values are zero for x(t), y(t), and their derivatives d dy t s Y s x t s X sdt dt

←→ ←→

Transfer function: Inverted transfer function:

( ) ( )( )

k k1 k 0 k 1

k kk 0 k 1

a s s dX s 1H sY s Kb s s c

− = =

−= = = ⋅

∑ ∏

The inverted transfer function H-1(s) has:• Poles equal to zeros of the transfer function H(s)• Zeros equal to poles of the transfer function H(s)

Inversion, Example 1

( ) ( )( )

( ) ( ) ( ) ( )

Y s s s 2 4Transfer function : H s s 2X s s 3 s 3dImpulse response: h t t 2 t 4e u tdt

+ −= = = − +

= δ − δ +

( ) ( )( )

1 t 2t

X s s 3 1 1Inverse transfer function : H sY s s s 2 s 1 s 2

Impulse response : h (t) u(t)e u(t)etwo poles : s 1, 2Inverse system is unstable.

− −

+= = = −

+ − − +

= −= −

( ) ( ) ( ) ( ) ( )2

d d dy t 3y t x t x t 2x tdt dt dt

+ = + −

Inversion, Example 2

( ) ( )( )

( ) ( ) ( ) ( )

X s (s 3)(s 2) 50Inverse Transfer function : H s s 8Y s s 7 s 7

dImpulse response: h t t 8 t 50e u tdt

− + −= = = + +

− −

= δ + δ +

( ) ( )( )

( ) ( ) ( )3t 2t

Y s 2 1 s 7Transfer function : H sX s s 3 s 2 (s 3)(s 2)

Impulse response: h t 2e u t e u t−

−= = − =

+ − + −

Magnitude Response (Amplitude Characteristics)Phase Response (Phase Characteristics)

( ) ( )( )

kk 1 kk 1

kk 1 k 1 k

s 1s cY s cFactorised transfer function : H s K K '

X s ss d 1d

−−

= = ⋅ = ⋅ − −

∏∏

∏ ∏

We apply for usually occuring systems M ≤ N

( )M M

k 1 k 1k k

Definition of magnitude response :

j jA 20log H j 20log K ' 20log 1 20log 1c d= =

ω ω= ω = + + − +∑ ∑

( ) ( )( )( )( )

k 1 k 1k k

Definition of phase response :

Im H jH j arctan K ' arctan arctan

Re H j c d= =

ω ω ωϕ = ∠ ω = = ∠ + − ω

∑ ∑

Bode Diagram of First-order Real Pole Factor

nNormalized cyclic frequency ω/ω

Positive real pole (unstable H)

Negative real pole

Bode Diagram of First-order Real Zero Factor

Negative real zero

Positive real zero

Example on a Transfer Function (1)

( ) ( )( ) ( )22

Y s s+2 s+2H sX s s 2s 1 s 1

= = =+ + +( ) ( ) ( ) ( ) ( )

d d dy t 2 y t y t x t 2x tdt dt dt

+ + = +

Frequency response: s = jω

( ) ( ) ( )( ) ( ) ( )2 2s j

Y j j + 2 j( / 2 )+1H j H s 2X j j 2 j 1 j 1= ω

ω ω ωω = = = =

ω ω + ω + ω +

( ) ( )

( )( ) ( ) ( )½ ½ ½2 2 2

A 20log H j

20log 2 20log / 2 1 20log 1 20log 1

ω = ω =

+ ω + − ω + − ω +

( ) ( ) ( ) ( )H 0 arctan / 2 arctan arctanϕ = ∠ ω = + −ω − ω − ω

Amplitude characteristics:

Phase characteristics:

( ) ( )

( )( ) ( ) ( )2 ½½2 2½

20log 2 20

Amplitude characteris

20log 1 20lolog / 2 1

tics : A 20 g H j

−+ ω + − ωω

Example on a Transfer Function(2)

110− 010 110 210 310

20log 2 10

( )A(dB)

( ) ( ) ( ) ( )arctan / 2 arctaH n a ctan0 r+ −ϕ = ∠ ω = ω − ω − ω

Example on a Transfer Function(3)Phase characteristics

110− 010 110 210 310

( )Hϕ = ∠ ω

Example on Transfer Function with Complex Poles

( ) ( ) ( ) ( )( ) 2s j

Y s 1H j H s H sX s j j2 1

= ωω = = = =

ω ω+ ζ + ω ω

( ) ( )( ) 2

Y s 1H sX s s s2 1

+ ζ + ω ω

( ) ( )

12 22 2

A 20log H j 20log 1 4 ω ω ω = ω = − − + ζ ω ω

( ) ( ) n2s j

2H j H s arctan

ω ζω ϕ = ∠ ω = = − ω − ω

Frequency response: s = jω

Bode Diagram of Second-order Pole Factor

20 ζ = 0.1

0ζ = 0.1

Complex pole-pair with negative real value

Bode Diagram of Second-order Zero Factor

ζ = 1.0

( )A ω

Complex zeros-pair in either left half or right half of the s-plane results in same amplitude characteristics

ζ = 0.1

1.0( )ϕ ω

Complex zeros-pair in left half of the s-plane

0Phase characteristics:

Complex zeros-pair in right half of the s-plane

ζ = -1.0- 0.5- 0.2

Minimum, Maximum and Mixed Phase Systems• Transfer functions H(s) with zeros in right and left half s-plane

with identical real parts results in:– Identical maginitude response | H(s) |– Different phase response ∠H(s)– Different impulse response h(t)

• Minimum phase systems:– Zeros in negative half s-plane only– Narrowest impulse respons h(t)– Inverse system H-1(s) containes poles in negative half s-plane only– Inverse system is stable

• Maximum phase systems:– Zeros in positive half s-plane only– Impulse respons h(t) is not the narrowest possible– Inverse system H-1(s) containes poles in positive half s-plane only– Inverse system is unstable

• Mixed phase systems:– Zeros in positive and negative half s-plane– Impulse respons h(t) is not the narrowest possible– Inverse system H-1(s) containes poles in positive half s-plane– Inverse system is unstable

Minimum and Maximum Phase Analogue Signals

Zero: s = αs=jω

s-plane

Minimum phase: zero in s < 0

When 0 ≤ ω ≤ ∞:- ϕ0 ≤ ϕ = ∠(jω − α) ≤ π/2

s-plane

Zero: s = α

s=jω ϕ

Maximum phase: zero in s > 0

When 0 ≤ ω ≤ ∞:π/2 ≤ ϕ = ∠(jω − α) ≤ π + ϕ0

Maximum and Minimum Phase Example on Analogue Signal

( ) ( )s 3Maximum phase : H

s 4 s 5−

( ) ( )s 3Minimum phase : H

s 4 s 5+

Maximum phase impulse response :h(t) 8e 7e− −= −

Minimum phase impulse response :h(t) 2e e− −= −

0 0.5 1 1.5 2 2.5-0.5

1Maximum phase

time (sec)

0 0.5 1 1.5 2 2.5-0.5

1Minimum phase

time (sec)

0 10 20 30 40 50-2

4Maximum phase

cyclic frequency (rad/sec)

0 10 20 30 40 50-1.5

0Minimum phase

∠H = arctan(ω/3) – arctan(ω/4) – arctan(ω/4)

∠H = π – arctan(ω/3) – arctan(ω/4) – arctan(ω/4)

Signal ProcessingPart 7.1 Digital Filters – Overview

and z-Transforms

Magnus Danielsen

Introductory Remarks

• FIR – filters:requires finite numbers of terms in a DFTF

• IIR – filters:requires infinite numbers of terms in a DFTF

• Equalization:requires inverted transfer functions

• Generalized method for construction of digital transfer function constructions:

z – transforms – a generalisation of DTFTz – transform propertiesz – transform applications

Elementary FIR Filter: FIR – filter

h[n]↔ H(ejΩ)

y[n]↔Y(ejΩ)x[n]↔X(ejΩ)

0 1 2 3 4 5 6 7-60

0 0.5 1 1.5 2 2.5 3 3.50

-30 -20 -10 0 10 20 30-0.05

hd[n]↔ Hd(ejΩ)

hd[n]=Ωc/π sinc[(Ω/π(n-M/2)]

Hd(ejΩ)=ejMΩ/2 for Ω≤Ωc=0 for Ω>Ωc

-40 -30 -20 -10 0 10 20 30 40-0.05

h[n] = hd[n] for n≤6= 0 for n>6

20logHd (dB)

Desired filter

Approximated filter of M+1=13th order

Infinite Impulse Response or IIR – Filters

• Transfer function is a DTFT corresponding to an infinite impulse response series

• Realisation requires feedback in the block diagram

• Best realized by generalization of DTFT by introduction of z – transform

• Method for filter construction will be based on bilinear transform, and converting of analogue transfer functions to digital transfer functions

z – Transform Definitionj

n n n jn n n

Define a complex number: z r e r cos jr sinSpecifically for z we obtain : z r e r cos n jr sin n

± ± ± ± Ω ± ±

= ⋅ = Ω + Ω

= ⋅ = Ω ± Ω

[ ] [ ] n

For a discrete signal: x n

we define z-transform: X z x n z∞

=−∞

= ∑±±±

[ ] ( ) [ ]z n

z transformation : x n X z = x n z∞

=−∞

− ←→ ∑

[ ] ( ) [ ]

DTFT j jn

DTFT transformation for comparison, replacing z by e :

x n X e = x n e

∞Ω − Ω

=−∞

←→ ∑

Inverse z – Transform [ ] ( ) [ ]z p

z-transform: x n X z = x p z∞

=−∞

←→ ∑

jz re Ω=

jdz rd je z jdΩ= Ω ⋅ = ⋅ Ωz-plane( )

[ ] [ ]

p n 1 n p 1

p pc c

n p j(n p)

Determine the integral X z z dz

x p z z dz x p z z jd

jx p r e d jx n 2

∞ ∞− − − −

=−∞ =−∞

∞− −

=−∞

= = ⋅ Ω

= Ω = ⋅ π

∑ ∑∫ ∫

∑ ∫

[ ] ( ) n 1

1Inverse z-transform: x n X z z dz2 j

−=π ∫

Convergence - ROC

( ) [ ] [ ] [ ]n n n

X z = x n z converges when x n z x n r∞ ∞ ∞

− − −

=−∞ =−∞ =−∞

≤ < ∞∑ ∑ ∑

Definition for region of convergencearea in z plane, where sum converges

== −ROC

[ ] [ ]n

Example with exponential discrete signal x n u n 1:

r ROC : z r∞

− −

= α α >

α < ∞ → = > α∑

Unit Circle in z – Plane is Transformed into the Frequency Axis for DTFT

unit circle :z r 1 and z e Ω= = =

jz re Ω=z-plane

( ) [ ]

X z = x n z

x n e X e

∞−

==−∞ =

∞− Ω Ω

=−∞

[ ] [ ]

( )( )

j j j j2

x n ...0 0 0 1 2 1 1 0 0 0...

z transform : X z z 2 z z

DTFT transform : X e e 2 e e

− −

Ω Ω − Ω − Ω

− = + − +

= + − +

Example comparing z – transform and DTFT

Difference Equation and z – Transform

[ ] ( ) [ ]

[ ] [ ] [ ] ( )0 0

(n n ) nz n0 0

x n X z = x n z

x n-n x n-n z x n z z X z

∞−

=−∞

∞ ∞− + −−

=−∞ =−∞

←→

←→ = = ⋅

∑ ∑

Time shift:

[ ] ( ) [ ] ( )[ ] [ ] ( ) ( )

x n X z y n Y z

a x n b y n a X z b Y z

←→ ←→

⋅ + ⋅ ←→ ⋅ + ⋅

Linearity:

k kk 0 k 0

− = −∑ ∑

Difference equation:

( ) ( )N M

k kk k

k 0 k 0

a Y z z b X z z− −

=∑ ∑z-transformed equation:

( ) ( )( )

b zY zH z

X z a z

= =∑

∑Transfer Function :

Transfer FunctionPoles and Zeros

x[n] y[n][ ] ( )zh n H z←→( )X z ( )Y z

( ) ( )( )

kk 01 M

0 1 M1 N

0 1 NM

0 k 1N

b zY zH z

X z a z

b b z ... b za a z ... a z

(1 c z )ba (1 d z )

− −

+ + +=

−= ⋅

Transfer Function :

Poles: dk where k=1 ... NZeros: ck where k=1 ... M

Causality and Anticausality – Two Examples

[ ] [ ] ( )zn n n1

Causal expon

1x n u n X z z ROC : z

ential signal:

1 z−

= α ←→ = α = > α− α∑

0 α Re(z)

[ ] [ ] ( )1

zn n n

n n 11 1

x n u n

Anticausal e

1 X z z

1 1z z ROC : z

xponential

gnal:−

=−∞

∞− −

− −=

= −α − − ←→ = −α

= −α = −α = < α− α − α

Two – sided SignalExample

[ ] [ ] [ ]

1ROC:z 11ROC: z2

1x n u n 1 u n2

1 1X z 1 11 z 1 z2

= − − − +

←→ = − + − −

( )( )

3z z2X z

11 z z2

1ROC : z 12

− = − −

0 ½ 1

Signal ProcessingPart 7.2 The z-Transform and its Properties

Magnus Danielsen

Properties of Region of Convergence (ROC)

ROC contains no polesROC = tot. z-plane, except z 0,z

Causal x[n]: X(z) x[n]z has only a pole in z 0

Anticausal x[n]: X(z) x[n]z has only a pole in z

•• = = ∞

− = =

− = = ∞

Finite duration signal x[n] :

Infinite duration si

-n -n -n

n n n 0

ROC = anular region in z-plane

X z x[n]z x[n]z x[n]z I I∞ − ∞

− +=−∞ =−∞ =

= = + = +∑ ∑ ∑

gnal x[n] :

ROC of I+

Causal signal part

ROC of I−

Anticausal signal part

ROC of X(z)

Two-sided signal

Examples of Two-sided ROC[ ] [ ] [ ] ( )

n n0n n

1 10 ROC: z2 1ROC: z

1 1x n u n 2 u n X z2 4

1 1 1 2z 2 z 12 4 1 2z 1 z4

∞− −

−−∞ <

= − − + ←→ =

− + = + + −∑ ∑ 1 1ROC : < z <

[ ] [ ] [ ] ( )n n

n nn n

1 10 01 1ROC: z ROC: z2 4

1 1y n u n 2 u n Y z2 4

1 1 1 2z 2 z 1 12 4 1 z 1 z2 4

∞ ∞− −

− −

= − + ←→ =

− + = + + −

∑ ∑ 1ROC : z >2

[ ] [ ] [ ] ( )n n

n n0 0n n

1 1ROC: z ROC: z2 4

1 1w n u n 2 u n W z2 4

1 1 1 2z 2 z2 4 1 2z 1 4z

− −

−∞ −∞ < <

= − − + − ←→ =

− + = + + − ∑ ∑ 1ROC : z <

Definition of ROC

ROC, named R is defined by r z r1 1 1ROC, named is defined by z

ROC of a right-sided signal is of the form z r causal signal

ROC of a left-sided signal is of the form z r anti causal signal

ROC of a two-sided signal is of the form: r z r non causal signal

< ⇒ −

< < ⇒ −

mixed causal and anti causal signal= −

Properties and Calculation Rules[ ] ( ) [ ] ( )

[ ] [ ] ( ) ( )

[ ] ( )

Linearity : ax n by n aX z bY z ROC: R R

1 1Time reversal : x n X ROC:z R

Time shift : x n n z X z ROC: R

zMultiplication exp. sequence: x n X ROC: R

Convolution

←→ ←→

+ ←→ +

− ←→

α ←→ α α

z zx yx n X z ROC R y n Y z ROC R

[ ] [ ] ( ) ( )

[ ] ( )

: x n y n X z Y z ROC: R R

dDifferentiation in z domain: nx n z X z ROC: Rdz

∗ ←→

←→−

Example: Application of Calculation Rules

[ ] [ ] [ ] [ ]

1ROC R R R : z 42

1 z 42x n w n y n X z 1 z 4z2 = < <

− −= ∗ ←→ = ⋅ − +

[ ] [ ] [ ] [ ] [ ] [ ] [ ]n n1 1w n n u n y n u n

− − ∗ = ⋅ = − x n = w n y n

[ ] [ ]

1ROC z2

ROC z 4

1 zu n 12 z2

1 z1 d z 2w n n u n z 1 12 dz z z2 2

1 4y n u n4 z 4

− ←→ ⇒ +

− − = ⋅ ←→− = + +

− = − ←→ −

Examples on Basic z – Transforms (1)[ ] [ ] ( )z n

10 ROC z 1

1 x n u n X z z1 z

∞−

= ←→ = =−∑Discrete step signal :

[ ] [ ] ( ) ( )zn n n

ROC z a

1 x n a u n X z a z1 az

∞−

= ←→ = =−∑Exponential signal :

[ ] [ ] ( )zROC total z plane

(the only function with ROC = total z plane):

x n n X z 1−

= δ ←→ =

Discrete impulse signal

[ ] [ ] ( )z k

ROC z 0x n n k X z z−

>= δ − ←→ =Discrete impulse signal :

Examples on Basic z – Transforms (2)

[ ] [ ] [ ]

( ) ( ) ( )( ) ( )

j n j n zn n n0

j n j nn n n nj j1 1

j j1 1

1 1x n a cos( n)u n a e a e u n2 2

1 1 1 1 1X z a e z a e z2 2 2 1 ae z 1 ae z

1 ae z 1 ae z12 1 ae z 1 ae z

Ω − Ω

∞ ∞Ω − Ω− −

Ω − Ω− −

= Ω = + ←→

= + = + − −

− + − ⋅= =

⋅ ⋅− −

∑ ∑-1

1 - acosΩ zX z1 - 2a cosΩ z + a

ROC z a>

Damped cosine function

[ ] [ ] [ ]

( ) ( ) ( )( ) ( )

j n j n zn n n0

j n j nn n n nj j1 1

j j1 1

1 1x n a sin( n)u n a e a e u n2 j 2 j

1 1 1 1 1X z a e z a e z2 j 2 j 2 j 1 ae z 1 ae z

1 ae z 1 ae z12 j 1 ae z 1 ae z

Ω − Ω

∞ ∞Ω − Ω− −

Ω − Ω− −

− Ω Ω− −

Ω − Ω− −

= Ω = − ←→

= − = − − −

− − − ⋅ ⋅= =

⋅ ⋅− −

∑ ∑-1

a sinΩ zX z1 - 2a cosΩ z

ROC z a>

-1 2 -2+ a z

Damped sine function

Inversion of z – TransformsPartial Fraction Method

( ) ( )( )

( )( )

1 Mk M Nkk 0 0 1 M

kN 1 Nk k 00 1 N

M N Nk k

k 1k 0 k 0 k

b zB z B zb b z ... b zX z f zA z a a z ... a z A za z

Af z (no multiple poles)1 d z

−− − −

−=− −

−−

−= =

+ + += = = = +

= +−

∑∑

∑ ∑

[ ] zn kk k k 1

ACausal solution ROC z d : A d u n1 d z−> ←→−

[ ] zn kk k k 1

AAnti causalsolution ROC z d : A d u n 11 d z−− < − − − ←→−

Example 1: Inversion of z – TransformsPartial Fraction Method

( )( ) ( )

( ) ( ) ( )

1 2 31 1

1 z zX z with ROC 1 z 211 z 1 2z 1 z2

A A AX z1 1 2z 1 z1 z2

− −

− − −

− −−

− += < < − − −

= + + − −−

[ ] [ ] [ ] [ ]n

n1x n u n 2 2 u n 1 2 u n2

= + ⋅ − − − ⋅

( ) ( ) ( )1

1 1 2 2A 1 z X z 12 1 2 2 1 2

Similarily : A 2 A 2−

− + = − = = − ⋅ − = = −

ROC of X(z)

½ 1 2

Poles and ROC

Example 2: Inversion of z – TransformsPartial Fraction Method

( )11 1

z 10z 4z 4X z2z 2z 4

z 1 10z 4z 4z2 1 z 2zz 4z 4z 10z 12 2z z 1z 5z 22z 32 2z z 1

z z2 2

1 32z 3 W z1 z 1 2z

− − −

− −

− − −

− −

−−

− −

−− −

− − +=

− −− − +

= ⋅− −− − +

= ⋅− −

− + + − + −

+ − −

= ⋅ − + + − − +

[ ] [ ] [ ] ( ) [ ] [ ]

[ ] [ ] [ ] ( ) [ ] [ ]n 1

w n 2 n 1 3 n 1 u n 13 1x n n n 1 1

u n 2 3 2 u n 2

u n+= −δ + δ + − − − − +

= − δ − + δ − − − − + ⋅ − −

⋅ − −

Poles and ROC: |z|<1

ROC of X(z)

Inversion of z – TransformsPower Series Expansion Method

2 z 1X z ROC z1 21 z2

+= ⋅ >

−Example1 :

[ ] [ ] [ ] [ ] [ ]

11 2 3

2 z 1X z 2 2z z z .....1 21 z2

1x n 2 n 2 n 1 n 2 n 3 .........2

−− − −

+= = + + + +

= δ + δ − + δ − + δ − +

( ) 2zX z e ROC z= ≠ ∞Example2 : ( )

z 2k n

k 0 n 2kn even

1 1X z e z znk! !

20 for n 0 or n odd

1x n for n 0 and n evenn !

∞−

= =− =−∞

= = =−

>= ≤ −

∑ ∑

Signal ProcessingPart 7.3 The z-Transform – Transfer Functions

Magnus Danielsen

Transfer Function• Definition of transfer function: H(z)=Y(z)/X(z), where Y(z)

and X(z) are output and input signals respectively• Initial conditions are zero• Causality – unilateral transforms• Stability – all poles are inside the unit circle of the z-plane• Minimum phase – all zeros are inside the unit circle of the

z-plane• Inverse system transfer function H-1(z) of H(z) such that

H-1(z)·H(z)=1• Frequency response is determined by poles and zeros• Decibel – definition:

– A=20 logH ,where H is a transfer function. – A=10 log(P2/P1), where P2 and P1 are the output and input powers

respectively.

Transfer Function

( ) ( )( )

b zY zH z

X z a z

= =∑

∑Transfer Function :

x[n] y[n][ ] ( )zh n H z←→( )X z ( )Y z

[ ] [ ] ( )

[ ] ( ) [ ] [ ] ( )

1 1 1x n u n X z ROC R z13 31 z3

1 3 1y n 3 1 u n u n Y z 13 1 z 1 z3

4 ROC R z 111 z 1 z3

−−

− −

= − ←→ = > +

= − + ←→ = + + −

= > + −

( )( ) ( )

y x y1 1

14 1 zY z 3H(z) ROC R R R : z 11X z 1 z 1 z3

− −

+ = = = >

Example on a transfer function

Transfer Function and Impulse Response

( ) ( ) ( )[ ] [ ] [ ] [ ] ( )[ ] [ ] [ ] [ ] [ ] [ ]

Y z H z X z

y n h n x n , where h n H z

x n n y n h n n h n

= ∗ ←→

= δ ⇒ = ∗ δ =

Z transform of output signal :

Output signal time function :

Impulse response h(t) :

[ ] ( )zx n X z←→ [ ] ( )zy n Y z←→[ ] ( )zh n H z←→

[ ] [ ] ( )( ) ( ) ( ) ( )[ ] [ ] [ ]

1Example on impulse response : H z1 az

x n n X z 1

Y z H z X z H z

y n h n a u n

−=+= δ ←→ =

Causal System • Causality requires that h[n]=0 for n<0• This requires rightsided transfer function

Stability for a Causal Signal• A pole |dk|<1 inside the unit circle, gives an

exponentially decaying term • A pole |dk|>1 outside the unit circle, gives

an exponentially increasing term • A pole |dk|=1 on the unit circle, gives an

complex sinusoid term (i.e. Constant ampl.)

Anti-causal System • Anti-causality requires that h[n]=0 for n≥0• This requires leftsided transfer function

Stability for a Anti-causal Signal• A pole |dk|<1 side the unit circle, gives an

exponentially decaying term for n → +∞• A pole |dk|>1 outside the unit circle, gives

an exponentially increasing term for n → -∞• A pole |dk|=1 on the unit circle, gives an

complex sinusoid term (i.e. Constant ampl.)

Example on Causality and Stability( )

[ ] [ ] [ ] ( ) [ ]

( ) [ ]

1j j1 14 4

j j4 4

n nj j n4 4

2 2 3H z1 2z1 0.9e z 1 0.9e z

The system shall be either or .

Poles z 0.9e z 0.9e z 2

Stable system : h n 2 0.9e u n 2 0.9e u n 3 2 u n 1

4 0.9 cos n u n 34

π π −−− −

π π−

= + ++

− −

= = = −

= ⋅ + ⋅ + − − −

π = + −

stabel causalSolution :

( ) [ ]

[ ] [ ] [ ] ( ) [ ]

( ) [ ] ( ) [ ]

n nj j n4 4

2 u n 1

Causal system : h n 2 0.9e u n 2 0.9e u n 3 2 u n

4 0.9 cos n u n 3 2 u n4

π π−

− −

= ⋅ + ⋅ + −

π = + −

Inverse Systemsx[n] y[n][ ] ( )zh n H z←→( )X z ( )Y z

[ ] ( )zinv invh n H z←→ x’[n]

( )X ' z

[ ] [ ][ ] [ ] [ ]( ) ( )

x ' n x n requires that

h n h n n

H z H z 1

∗ = δ

( ) ( )( ) ( )( ) ( )

1H zH z

Poles in H z are converted to zeros in H z

Zeros in H z are converted to poles in H z

Example on a Stable and Causal Inverse System

1 1 1y[n] y[n 1] y[n 2] x[n] x[n 1] x[n 2]4 4 8

− − + − = + − − −Difference equation :

x[n] y[n][ ] ( )zh n H z←→( )X z ( )Y z

[ ] ( )zinv invh n H z←→ x’[n]

( )X ' z

( ) ( )( )

( ) ( )

1 2 1 1

1 2 1 2

1 1 1 11 z z (1 z )(1 z )Y z 4 8 4 2H z 1 1X z 1 z z (1 z )4 2

1 1 1z and z z (double pole)4 2 2

− − − −

− − −

+ − − += = =

− + −

= = − =

Transfer function :

Zeros for H z : Poles for H z :

( ) ( )( )

( ) ( )

1(1 z )Y z 2H z 1 1X z (1 z )(1 z )4 2

1 1 1z (double zero) z and z2 4 2

Both poles are inside unit circle stable inverted systemBoth zeroes are inside

− −

−= =

= = = −

inv inv

Inverse Transfer function :

Zeros for H z : Poles for H z :

unit circle minimum phase inverted system⇒

Example on Multipath Communication System

y[n] x[n] ax[n k]= + −Difference equation :

( ) ( ) ( )

r phase(a)1 1 jkk k

H z 1 az

z a a e where r 1,3,...,2k 1

= − = = −

Transfer function:

Zeros for H z : No polesfor H z

( ) ( )( )

( ) ( ) ( )

r phase(a)1 1 jkk k

Y z 1H zX z 1 az

z a a e where r 1,3,..., 2k 1

All poles are inside unit circle if a <1 stable inverted system

= =−

= − = = −

inv inv

Inverse Transfer function :

Poles for H z : No zeros for H z

Sampling time = TsTime difference between direct and reflected signal: ∆T=kTs

Direct signal

Reflected signal

Transmitter Receiver

Mountain

Unilateral z – Transform used for Causal Systems

[ ] [ ]

Definition : x n is defined for n

unilateral z transform : x n x n z ,

and bilateral z transform : x n x n z

are identical for causal signals, i.e. if x n 0 for n 0

∞−

=−∞

− ∞ < < ∞

− ←→

[ ] [ ]uz n

x n x n z∞

←→∑ [ ] [ ] [ ] [ ]uz n 1 n

n 0 n 0

x n-1 x n-1 z x -1 z x n z∞ ∞

− − −

←→ = +∑ ∑

[ ] [ ] [ ] [ ] [ ] [ ]uz n 1 k 1 k n

n 0 n 0

x n-k x n-k z x -k x -k+1 z ...x -1 z z x n z∞ ∞

− − − + − −

←→ = + + +∑ ∑

Difference Equation with Initial Condition

k kk 0 k 0

− = −∑ ∑Difference equation:

z-transformed equation:

( ) ( ) ( ) ( ) ( )A z Y z C z B z X z+ =

A z a z−

= ∑ ( )M

B z b z−

= ∑ ( ) [ ]N 1 N

m 0 k m 1

C z a y k m z−

= − +∑ ∑

Initial conditions for y[n]: y[-1], y[-2] , y[-3] , y[-4] , .... , y[-N+1]

Initial conditions for x[n]: 0, 0, 0, ... i.e. X[n] is taken to be causal

Signal ProcessingPart 7.4 The z-Transform – Transfer Functions Frequency Response and Block Diagrammes

Magnus Danielsen

Frequency Response

( ) ( )j

kj N M k 1

(e c )with H z K e

(e d )Ω

Ω − ==

−= ⋅

∏jΩTransfer Function z = e :

( ) ( )( )

1 Mk0 1 Mk 0

N 1 Nk 0 1 N

k kN Mk 1 k 1

k kk 1 k 1

b zY z b b z ... b zH zX z a a z ... a za z

(1 c z ) (z c )K K z

(1 d z ) (z d )

−− −

=− −

−= =

+ + += = =

− −= ⋅ = ⋅

− −

∏ ∏

Transfer Function :

jΩegjΩIllustration of the factor : e - g

jΩe - g

Example: Amplitude and Phase Characteristics

-4 -3 -2 -1 0 1 2 3 4-2.5

-4 -3 -2 -1 0 1 2 3 40

( ) ( )1 1

1 2j j1 14 4

1 z 1 zH z1 0.9 2z 0.81z1 0.9e z 1 0.9e z

− −

π π − −−− −

+ += = − +− −

Transfer function :

( ) ( )

z e j 2 j

1 eH z

1 0.9 2e 0.81e

2 2cos

1 0.9 2 cos 0.81cos2 0.9 2 sin 0.81sin 2

− Ω

= − Ω − Ω

− Ω + Ω + Ω + Ω

Amplitude characteristic

( )( )

sin( ) sin( )sin 4 4arctan arctan arctan1 cos 1 cos 1 cos

4 4sin 0.9 2 sin 0.81sin 2arctan arctan1 cos 1 0.9 2 cos 0.81cos2

Ω=∠ =

π π−Ω + −Ω −−Ω

− −π π+ −Ω + −Ω + + −Ω −

−Ω Ω + Ω

= −+ −Ω − Ω + Ω

Phase characteristic

Minimum, Maximum and Mixed Phase Systems

• Transfer functions H(z) with zeros inside the unit-circle and outside the unitcircle in the z-plane with reciprocal values result in:

– Identical maginitude response | H(z) |– Different phase response ∠H(z)– Different impulse response h[n]

• Minimum phase systems:– Zeros inside the unit circle in the z-plane only– Narrowest impulse respons h[n]– Inverse system H-1(z) containes poles inside the unit circle in the z-plane only– Inverse system is stable

• Maximum phase systems:– Zeros outside the unit circle in the z-plane only– Impulse respons h[n] is not the narrowest possible– Inverse system H-1(s) containes poles outside the unit circle in the z-plane only– Inverse system is unstable

• Mixed phase systems:– Zeros inside and outside the unit circle in the z-plane– Impulse respons h[n] is not the narrowest possible– Inverse system H-1(s) containes poles outside the unit circle in the z-plane– Inverse system is unstable

Minimum and Maximum Phase Discrete Signals( ) 0j1H z 1 az Zero : z a | a | e ϕ−= − = =

Minimum phase: Zero z=a inside unit circle |z| < 1

When – π ≤ Ω ≤ π:- Arcsin|a| ≤ ϕ = ∠(1-az-1) ≤ Arcsin|a|

z=e-jΩ

z-plane

z=a- Ω

z=ae-jΩ

1–ae-jΩ

Maximum phase: Zero z=a outside unit circle |z| >1

When – π ≤ Ω ≤ π:– π ≤ ϕ = ∠(1-az-1) ≤ π

z=e-jΩ

z-planeϕ

z=a- Ω

z=ae-jΩ

1–ae-jΩ

Maximum and Minimum Phase Example on Discrete Signal

Maximum phase :1 2zH

1 11 z 1 z3 4

− −

+= + +

Minimum phase :2 zH

1 11 z 1 z3 4

− −

+= + +

-4 -2 0 2 4-4

4Maximum phase

-4 -2 0 2 4-4

4Minimum phase

-2 0 2 4 6 8 1 0-1 .5

1M a xim um p ha se

-2 0 2 4 6 8 1 0-0 .5

2M inim um p ha se

[ ] [ ] [ ]n n

Maximum phase impulse response :

1 1h n 20 u n 21 u n3 4

= − +

[ ] [ ] [ ]n n

Minimum phase impulse response :

1 1h n 4 u n 6 u n3 4

= − +

Difference EquationsN M

k kk 0 k 0

− = −∑ ∑

k kk 0 k 1

y[n] b x[n k] a y[n k]= =

= − − −∑ ∑

Recursive formula for y[n]

( ) ( ) ( )M N

k kk k

k 0 k 1

Y z b z X z a z Y z− −

= −∑ ∑

z-transform Y(z) for y[n]

Block Diagram – Discrete 2nd ordery[n] = b0x[n] + b1x[n-1] + b2x[n-2] - a1y[n-1] – a2y[n-2]

Y(z) = b0X(z) + b1z-1X(z) + b2z-2X(z) - a1z-1 Y(z) – a2z-2Y(z)

W(z)X(z)

z-1 z-1

Y(z)b0+

( ) ( )( )

Y zH z

Block Diagram – Discrete N’th ordery[n] = b0x[n] + b1x[n-1] + b2x[n-2]+ ...+ bMx[n-M] – a1y[n-1] – a2y[n-2] – .... – aNy[n-N]

wY(z) = b0X(z) + b1z-1X(z) + b2z-2X(z)+...+ bMX(z) – a1z-1 Y(z) – a2z-2Y(z) –...– aNz-2Y(z)

( ) ( )( )

b zY zTransfer function : H z

X z a z

= =∑

Y(z)b0+

Cascade-form Implementation

Y(z)b’0+

-a’2

-a’1

( ) ( )( )

( )( )

( )( ) ( ) ( )1 2

Y z W z Y zH z H z H z

X z X z W z= = ⋅ = ⋅

Example: Cascade-form Implementation

( ) ( )( )( )1 1

1j j1 1

2j j1 18 84 4

1 jz 1 jzH

1 11 e z 1 e z

3 31 e z 1 e z42 2 4

H z− −

π π−− − −−−

− −

( ) ( )( )

W z1 zH z1 X Z1 cos z z

− −

π − +

( ) ( ) ( )( )

1 z Y zH z

3 9 W z1 cos z z2 8 16

− −

π − +

cos(π/4)

Y(z)1+

3/4⋅cos(π/8)

Parallel-form Implementation+

Y’(z)b’0+

-a’2

-a’1

Y’’(z)b0+

X(z) + Y(z)=Y’(z)+Y’’(z)=[H’(z)+H’’(z)]X(z)

Signal ProcessingPart 8.1 Analogue Filters

Magnus Danielsen

Filter Types• Ideal filters:

– Constant amplitude in passband

– Linearly varying phase versus frequency

• Low-pass filters (LP)• High-pass filters (HP)• Band-pass filters (BP)• Band-stop filters (BS)• Characteristics (amplitude and

phase):– Butterworth filters– Chebyshev filters

• Frequency transformations– LP to HP transformation– LP to BP transformation– LP to BS transformation

• Analogue filters– Passive– Active filters– Electronic circuits designed

filters• Digital filters

– FIR– IIR– Processor based designed

filters

Distortion free Ideal Transmission of Analogue Signal

0j ts j

Y(s)H( j ) H s CeX(s)

h(t) C t t

− ω= ω

ω = = =

= δ −

( ) ( )Lh t H s←→ ( ) 0y t Cx(t t )= −( )x t0stY(s) CX(s)e−=

H(jω)

Amplitude characteristics

ϕ=∠ H(jω)

dϕ/d ω= - t0

Phase characteristics

Ideal transmission: constant amplitude amplification C, and group delay t0 for all frequencies composing the input signal x(t)

Ideal Low-pass Filters( ) ( )Lh t H s←→ ( )y t( )x t

e forH( j )

− ω ω ≤ ωω = ω > ω( )

j ( t t )j t j t

c 0 c c0

1 1 eh(t) e e d2 2 j(t t )

sin (t t )sinc (t t )

(t t )

ωω ω −− ω ω

−ω −ω

= ω = π π −

ω − ω ω = = − π − π π

H(jω)

− ωc ωc

ϕ=∠ H(jω)ω

− ωc ωc

-3 0 -2 0 -1 0 0 1 0 2 0 3 0-0 .1

-0 .0 5

0 .0 5

0 .1 5

0 .2 5

0 .3 5

Transmission of Rectangular Time Pulse Gibbs Phenomenon

T1 for t2x(t)T0 for t2

≤= >

( ) ( )Lh t H s←→ ( )y t( )x t

( )c 0 c c0

sin (t t )h(t) sinc (t t )

(t t )ω − ω ω = = − π − π π

( ) ( ) ( )0

0 0c 0 c 0

sin (t t )y(t) x h t d d

(t t )

1 T TSi (t t ) Si (t t )2 2

−∞ −

ω − − τ= τ − τ τ = τ

π − − τ

= ω − + − ω − − π

∫ ∫

sinSi(u) d

λ= λ

Rectangular input pulse: Ideal low-pass filter impulse response :

Output pulse response:

Sine Integral Function Si(u)

-40 -30 -20 -10 0 10 20 30 40-2

2Sine integral

u (radian)

sinSi(u) d

λ= λ

maxSi(u) 1.1790

≅ ⋅

/ 2−π

Gibbs PhenomenonRectangular Pulse filtred in Ideal Bandpass Filter

-0.5 0 0.5 1 1.5-0.2

Time (sec)

l ωcT0 =200

-0.5 0 0.5 1 1.5-0.2

Time (sec)

l ωcT0 =20

-0.5 0 0.5 1 1.5-0.2

Time (sec)

-0.5 0 0.5 1 1.5-0.2

Time (sec)

l ωcT0 =4ωcT0 = 2

maxy 1.09

maxy maxy

Design of Filters

• Prescription of frequency response of amplitude and phase characteristics

• Function selected must be causal and stable

• Realization of approximating transfer function by a physical system

• Analogue approach

• Analogue to digital approach

• Direct digital approach

0 ωp ωs

Passband Stoppband

Transition band

|H(jω)|2

1 ε1=

Fundamental Filter Parameters

Definition of filter passband parameters:• Passband tolerance parameter ε• Passband maximum attenuation αp = − 10⋅log(1 − ε) dB• Passband cutoff cyclic frequency ωp radian/sec• Stopband cutoff frequency fp = ωp/(2π) Hz

Definition of filter passband parameters:• Stopband tolerance parameter δ• Stopband minimum attenuation αs = − 10⋅ log(δ) dB• Stopband cutoff cyclic frequency ωs radian/sec• Stopband cutoff frequency fs Hz

• Maximally flat responsedkH(jω)/dωk=0 for ω=0 and ω=∞ :

– Butterworth filter characteristics

• Equiripple magnitude response– In passband, Chebyshev I filter characteristics– In stopband, Chebyshev II filter characteristics– In both passband and stopband, Elliptic filter

characteristics

Approximating Functions

Magnitude Response Description for Low-pass Filters

( ) ( )2 22s j

Transfer function for analogue filter: H s1Magnitude response: H s H j

1 F ( )= ω= ω =

Functions to be selected for filter design:

Butterworth filters F( ) maximally flat filter

Chebyshev I filters F( ) T equiripple filter in passband:

1Chebyshev II filters F( )T

ωω = ω

ωω = γ ω

ω =ω

T is the Chebyshev polynomium equiripple filter in stopband :

Elliptic filters F( ) R equiripple filter in both passband and stopband: R is the elliptic function

ω = ω

Digital Filters used for Analogue Signals

• Signal is samples with sampling time Ts• Transfer function is described by

H(ejΩ), where -π < Ω < π• Normalized cyclic frequency = Ω• Cyclic frequency ω = Ω/Ts• Transfer function in cyclic frequency

domain:H = H(ejωTs) where -π/Ts< ω < π/Ts

• H(ejωTs) is matemathically periodic in ω

Signal ProcessingPart 8.2 Analogue Butterworth Filters

Magnus Danielsen

Butterworth Filters( ) 2

1H j K is an integer number

ω = ω

0 ωp ωs

Passband Stoppband

Transition band

H(jω)2

ε ω = ω − ε

− δ ω = ω δ

αp = − 10⋅log(1 − ε) dBαs = − 10⋅ log(δ) dB

Butterworth Transfer Function( ) ( ) ( )

( ) ( )

1H s H s H jj1j

1H s H ss1

= ω− = ω =

ω+ ω

( ) ( )1 (2k 1)j

2K 2Kc c

1 (2k 1)j ( )2 2K

Poles of H s H s :

s j ( 1) j e

e , where k 1,2,...,2K

−π +

= ω − = ω

= ω =

Stability for H(s) requires the poles of H(s) to be placed in left half-plane

Poles of H(s) H(-s) are distributed with equal angular distance on a circle with radius = ω :

Butterworth Transfer Function( ) ( ) 2K

1H s H ss1

( ) ( )1 (2k 1)j

2K 2Kc c

1 (2k 1)j ( )2 2K

Poles of H s H s :

s j ( 1) j e

e ,where k 0,1,...,2K 1

= ω − = ω

=ω = −

K= 4Poles for H(s) Poles for H(-s)

K=3Poles for H(s) Poles for H(-s)

−ω= ω

ω= ω

Prototype transfer functions are defined by ωc=1

Design of Buttereworth LP-FiltersDesign a Butterworth (prototype) LP-filter with ωc=1, and K=3

Poles: s= −1,

s=e-j2π/3=-½+j½√3 and s=e-j2π/3=-½-j½√3

( )( )

1 1H ss 2s 2s 11 3 1 3s 1 s s

2 2 2 2

= =+ + +

+ + − + +

Change of bandwidth from ωc = 1 to ωc = ωc :

Substitution of s by s/ωc results in:

( ) 3 2

c c c c c c

1 1H ss s 1 3 s 1 3 s s s1 2 2 12 2 2 2

= = + + − + + + + + ω ω ω ω ω ω

Butterworth Filter Prototype Transfer Functions (ωc=1)

5 4 3 2

6 5 4 3 2

1K 1: H ss 1

1K 2 : H ss 2s 1

1K 3 : H ss 2s 2s 1

1K 4 : H ss 2.6131s 3.4142s 2.6131s 1

1K 5 : H ss 3.2361s 5.2361s 5.2361s 3.2361s 1

1K 6 : H ss 3.8637s 7.4641s 9.1416s 7.4641s 3.8637s 1

= =+ +

= =+ + +

= =+ + + +

= =+ + + + +

= =+ + + + + +

Butterworth Polynomials (denominator of H(s))

General FormulasEven polynomials of K’th order:

Examples on odd Butterworth polynomials:B1 = s+1B3 = (s+1) ⋅ (s2 + 2s cos(4/6 π) + 1)B5 = (s+1) ⋅ (s2 – 2 s cos(6/10 π) + 1) ⋅ (s2 –2 s cos(8/10 π) + 1)B7 = (s+1) ⋅ (s2 – 2 s cos(8/14 π) + 1) ⋅ (s2 –2 s cos(10/14 π) + 1) ⋅ (s2 –2 s cos(12/14 π) + 1)

K / 22

2n K 1B (s) s 2s cos 12K=

+ − = − π + ∏

( )( )K 1 / 2

2n K 1B (s) s 1 s 2scos 12K

+ − = + − π + ∏

Examples on even Butterworth polynomials:B2 = s2 – 2 s cos(3/4 π) + 1B4 = (s2 – 2 s cos(5/8 π) + 1) ⋅ (s2 –2 s cos(7/8 π) + 1)B6 = (s2 – 2 s cos(7/12 π) + 1) ⋅ (s2 –2 s cos(9/12 π) + 1) ⋅ (s2 –2 s cos(11/12 π) + 1)B16 = (s2 – 2 s cos(17/32 π) + 1) ⋅ (s2 –2 s cos(19/32 π) + 1) ⋅ (s2 –2 s cos(21/32 π) + 1) ⋅

(s2 – 2 s cos(23/32 π) + 1) ⋅ (s2 –2 s cos(25/32 π) + 1) ⋅ (s2 –2 s cos(27/32 π) + 1) ⋅(s2 – 2 s cos(29/32 π) + 1) ⋅ (s2 –2 s cos(31/32 π) + 1)

Odd polynomials of K’th order:

Butterworth Amplitude Characteristics for LP-filter

Slope = K·20 dB/decade

ω/ωc

Butterworth Phase Characteristics for LP-filter

ω/ωc

Design of Butterworth LP-filters( ) 2

1Amplitude characteristics : A 10log H j 10log

= ω = ω+ ω

Parameters K, ,Requirements: passband 0 A 20 log(1 ) dB

stopband A 20 log dB

ε δ≤ ω ≤ ω ≤ −α = − ε

ω ≤ ω ≤ ∞ ≥ −α = δ

( ) ( )( )( )

ps 0.10.1

log 10 1 / 10 1K

2 log /

αα − −≥

0.1 2K10 1α

ωω =

−( )s

10.1 2K

s c 10 1αω = ω −

Example: Butterworth LP filter (1)Given: Passband attenuation: αp = 2 dB for 0 < f < fp = 8 kHz

Stopband attenuation: αp = 15 dB for fs = 8 kHz < f < ∞

( ) ( )( )( )

ps 0.10.1 1.5 0.2

log 10 1 / 10 1 log 10 1 / 10 1K 4.88 5

2 log 12 / 82 log /

αα − − − −≥ = = ≅

⋅⋅ ω ω

( ) 5 4 3 2

3 3 3 3 3

1H ss s s s s3.2361 5.2361 5.2361 3.2361 1

53 10 53 10 53 10 53 10 53 10

= + + + + + ⋅ ⋅ ⋅ ⋅ ⋅

0.1 2K

ff 8.44 kHz

10 1α

= =−

10.1 2K

s cf f 10 1α= −

3 3c p2 f 2 8.44 10 53.0 10 rad / secω = π ⋅ = π ⋅ ⋅ = ⋅

Example: Butterworth LP filter (2)

Frequency (kHz)

Slope =20 K dB/dec- 120 dB/dec

K⋅90o=450o

K=5 fc = 8.44 kHz

Signal ProcessingPart 8.3 Analogue Chebyshev Filters and

Frequency Transformations of Filters

Magnus Danielsen

Chebyshev-I Filters( ) 2

1H j K is an integer number1 T

ω = ω

+ γ ω

0 ωp ωs

Passband Stoppband

Transition band

H(jω)

------ for odd K

for even K

Chebyshev Polynomials( )( )( )( )( )( )

T 8 8 1

T 16 20 5

ω = ω

ω = ω −

ω = ω − ω

ω = ω − ω −

ω = ω − ω − ω

( ) ( )( ) ( )

( ) ( ) ( )

K 1 K K 1

T cos K cos 1

T cosh K cosh 1

Recursion (generation) formula:T 2 T T

ω = ω ω ≤

ω = ω ω ≥

ω = ω ω − ω

Design of Chebyshev-I LP filters

Parameters K, ,Requirements: passband A 20 log dB

stopband A 20 log dB

ε δ− ≤ α = − ε

− ≥ α = − δ

( ) ( )( )( )

ps 0.10.11

cosh 10 1 / 10 1K

cosh /

αα−

− −≥

1H j K is an integer number1 T

ω = ω

+ γ ω

( ) ( )2 2

1H s H ss1 T

+ γ ω

Chebyshev Polynomials and Chebyshev-I Amplitude Characteristics

-1.5 -1 -0.5 0 0.5 1 1.5-10

Chebyshev polynomium

TK(ω/ωp)

K=1,2,3,4,5, and 6

ω/ωc

-1.5 -1 -0.5 0 0.5 1 1.5-20

ω/ωc

A(ω/ωp)=

-10 log (1+γ2TK2(ω/ωp))

Chebyshev ampl.char.

K=1,2,3,4,5, and 6

ε=0.106 , and γ=0.5

Poles of Chebyshev-I LP-filter

( )1 11 1 1sinh sinh or sinh K sinhK

− − η = = η γ γ

Definition of parameter :

( ) ( )( ) ( ) ( )2 2 1

Poles of H s H s will be found from :

T 1 and T cos K cos−−

γ ω = − ω = ω

( ) ( )1 2

r r rc

s jcos(sin j ) sin j 1 cos

(2r 1)where where r 1,..., 2K2K

= − η+ θ = −η θ − + η θω

−θ = π =

Poles for H s H s

( ) ( )2 2

1H s H ss1 T

+ γ ω

Stability for H(s) requires the poles of H(s) to be placed in left half-plane

⋅ −2

Poles of H(s) H( s) are placed on an ellipse

with the axes ηand 1 + η :

21+ η

Chebyshev-I Prototype LP Transfer Function( )

1H s Cs [ sin j 1 cos ]

C must be determined from |H(j )| for ω in passband =

=+ η θ + + η θ

K = 3 γ = 0.5 η = 0.5 ωc = 1

Poles for H(s) Poles for H(-s)

1 0.97 j4

− += −ω − −

1 0.97 j4

−= ω +

( )cPrototype LP transfer function 1:

0.5H s1 1 1s s 0.97j s 0.97j2 4 4

= + + − + +

Prototype transfer functions are defined by ωc=1

( )H s 1 for s 0 C 0.5= = ⇒ =

Chebyshev I Transfer FunctionK=4 γ=0.5 ωc=1

0.1411 j0.98470.3407 j0.4079s0.3407 j0.40790.1411 j0.9847

− +− += − −ω − −

( ) ( ) ( ) ( ) ( )

cPr ototypeLP transfer function1H s C

s 0.1411 j0.9847 s 0.3407 j0.4079 s 0.3407 j0.4079 s 0.1411 j0.9847

=+ + + + + − + −

1 21 1sinh sinh 0.3688 1 1.0658K

− η = = + η = γ

Poles for H(s) Poles for H(-s)

( )H s 1 for s 0 C 0.2500= = ⇒ =

0.1411 j0.98470.3407 j0.4079s0.3407 j0.40790.1411 j0.9847

− −− −= − +ω − +

Example: Chebyshev I LP Filter Given: Passband attenuation: αp = 2 dB for 0 < f < fp = 8 kHz

Stopband attenuation: αp = 15 dB for fs = 12 kHz < f < ∞

( ) ( )( )

ps 0.10.11 1 1.5 0.2

cosh 10 1 / 10 1 cosh 10 1 / 10 1K 2.77 3

cosh 12 / 8cosh /

αα− −

−−

− − − −≥ = = ≅

c pf f 8 kHz= =

2 f 2 8 10

50 10 rad / sec

ω = π ⋅ = π ⋅ ⋅

1 ss p

Resulting stopband frequency:

f2.77f ' f cosh cosh 11.373 f

− = =

1LP transfer function 50 : H s Cs s 1 s 11 0.97j 0.97j

50 10 50 10 4 50 10 4

ω = = ⋅ + + − + + ⋅ ⋅ ⋅

Example: Chebyshev I LP FilterK=3 fc = 8 kHz

Frequency (kHz)

K⋅90o=270o

Example: Chebyshev I LP Filter K=4 fc = 8 kHz

Frequency (kHz)

K⋅90o=360o

Frequency Transformations: Low-pass Prototype to Low-pass

(shift of cut-off frequency)

sLow-pass prototype to low-pass transformation: s

1 1ss d d

→− −

( ) ( )

LP ,prot 2

LP 2 2 2c c c

sExample : Butterworth filter : s

1H (s)s 1 s s 1

1H (s)s s ss s s1 1

= →+ + +

+ ω + ω + ω + + + ω ω ω

Frequency Transformations: Low-pass Prototype to High-pass

Low-pass to high-pass transformation: ss

s1 1 d

s d d ss d

−→ =

ω ω− − −

( ) ( ) ( ) ( )

LP,prot HP2 2

Example : Butterworth filter : s , 1s

1 1 sH (s) H (s)1 1 1s 1 s s 1 s 1 s s 11 1s s s

ω→ ω =

= → = = + + + + + ++ + +

Frequency Transformations:Low-pass Prototype to Band-pass

2 20 0 0

s sLow-pass to band-pass transformation: sBs B s

+ω ω ω→ = + ω

( ) ( )

1 20 0

1,22 2

1 1 Bss d s p s ps d

½Bd ½Bdp

½Bd ½Bd

→ =− − − ω ω+ − ω

+ − ω= − − ω

Example: Butterworth Band-pass Filter0 0

sTransformation : sB s ω ω

→ + ω

( ) ( )LP,prot 2

2c LP,prototype

1Prototype filter: H (s)s 1 s s 1

with 3-dB cut-off frequency 1, where 10 log(|H | ) 3dB

=+ + +

ω = ω = ⋅ =

0 0 0 0 0 0

1,2 ( )

Band pass filter :1H (s)

s s s1 1B s B s B s

swith 3-dB cut-off frequencies found from putting =1 B s

giving 4 solutions ½ B ½ B

= ω ω ω ω ω ω + + + + + + ω ω ω

ω ω+ ω

ω = ± 2 20

, two of them positive.

The 3 dB bandwidth B

− ω − ω =

112 20 0

s B sLow-pass to stop-band transformation: sBs s

−− +ω ω

→ = + ω ω

( ) ( )( ) ( )

0 1 20 01

1 20 00

ss s z s z1 1

s d Bd Bd s p s ps 1B s d B s ds

ω+ ω − −ω ω → = − = −− − − ω ω ω + −+ − ωω ω

Frequency Transformations: Low-pass Prototype to Stop-band

21 1 20

1,221 1 2

Poles :

½Bd ½Bdp

½Bd ½Bd

− −

+ − ω= − − ω

Zeros :j

= − ω

LP and BP Butterworth FilterBP

0 0 0 0 0 0

1H (s)s s s1 1

B s B s B s

= ω ω ω ω ω ω + + + + + + ω ω ω

1H (s)s s s1 1

= + + + ω ω ω

LP: K=3 fc=ω/2π= 8kHz BP : K=3 fc=ω/2π= 8kHz ∆f = =B/2π 2kHz

Frequency (kHz) Frequency (kHz)

LP and BS Butterworth Filter

LP: K=3 fc=ω/2π= 8kHz

1H (s)s s s1 1

= + + + ω ω ω

BS : K=3 f0=ω/2π= 8kHz ∆f = =B/2π 2kHz

BS 21 1 1

0 0 0 0 0 0

1H (s)B s B s B s1 1

− − −=

ω ω ω + + + + + + ω ω ω ω ω ω

Frequency (kHz) Frequency (kHz)

Signal ProcessingPart 8.4 Analogue Filters – Practical Constructions

Magnus Danielsen

Passive Butterworth LP-Filter of Third OrderV1

1 01 1 1

1 1 1sC EsL R sL VNode equations : R

V1 1 1 0sCsL sL R

+ + − = − + +

1 21 1 1 1

0 1 1 2 3 2

1 2 1 1 1 2 1 1 2

3 2 2 31 1 2 c c c

E1sL RV

1 1 1 1 1 1sC sCsL R sL R sL sL

V 1 1E RL C C 1 1 1 1 2s s s

RC RC L C L C L C C R1 1

RL C C s 2s 2s1K

s s s2 2

+ + + + − − −

+ + + + +

= ⋅+ ω + ω + ω

+ + ω ω ω 1+

2 3c c c 3

1 2 1 1 1 2 1 1 2 1 1 2 c

1 1 1 1 2 1 12 2 KRC RC L C L C L C C R RL C C 2

ω = + ω = + ω = = =ω

c c 1 1 2Example : R 50 f 159 Hz 1000 rad/sec L 50 mH C 68.3uF C 11.7 uF= Ω = ω = = = =

Passive Butterworth LP-Filter of Third Order (2)

Butterworth LP-filter Characteristics of 3rd Order

Passive Butterworth BP-Filter of 3rd Order

20 0 c c 0

Ls 1 1Inductances: sL L s L sL'B s B B s sC'

B 1Serial coupling of L' and C' L' L C'B L L'

ω ω ω ω ω → ω + = + = + ⇒ ω

ω= = =

ω ω ω

out 3 2

1LP-prototype filter: Vs 2s 2s 1

=+ + +

sPrototype Low-pass to band-pass transformation: sB s ω ω

→ + ω

0 0 0 0c

Low-pass to band-pass transformation cutoff frequency = :

s s sor sB s B s

ω ω ω ω→ + →ω + ω ω ω

20 0 c c 0

Cs 1 1Capitances: sC C s C sC''B s B B s sL''

B 1Parallel coupling of C'' and L'' C'' C L''B C C''

ω ω ω ω ω → ω + = + = + ⇒ ω

ω= = =

ω ω ω

Center frequency:f0 = ω0/(2π)

Band width:∆f = B/(2π)

Passive Butterworth BP Filter of 3rd Order

Phase characteristics ∆f = 800 Hzf0 = 1000Hz

Active LP-filter – Single Real Pole

11out 1

in 3 3 1 1

1 sCRV R 1H(s)

V R R 1 sR C

1Real pole: sR C

+ = = − = −

Active LP-filter – Single Real PoleFrequency Response

Magnitude response

Phase response

Example: LP to BP Transformation

1Prototype LP-filter: H(s)s 1

1 B 0.1s 1s ( ) 10(s )

B s s1 0.1sBandpass filter : H(s) 1 s 0.1s 110(s ) 1

ω = =ω ω

→ + = +ω

= =+ ++ +

Poles : s 0.05 j0.998 0.05 jZero : s 0

= − ± ≅ − ±=

Transfer Function of BP-filter

1 0- 1

- 1 . 5

- 0 . 5

0.1sH(s)s 0.1s 1

Active BP-filter Two Complex Poles, and a Zero

11 1out

2 131 1

1 1 1 1 1 1

1 1sCR sLVH(s)

V RL s

LR s L C s 1R

1 1 1s j2R C L C 2R C

+ + = = −

= − ⋅+ +

= − ± −

Transfer function :

Complex poles :

Zero :

Active BP-filter – Two Complex Poles, and a ZeroMagnitude and Phase

Center frequency:

1 1f 1007Hz2 L C

2 131 1

H(s)L s

LR s L C s 1R

− ⋅+ +

Phase Characteristic

Amplitude Characterlistic

Signal ProcessingPart 8.5 Digital FIR Filter

Magnus Danielsen

FIR-Filter Properties and Applications

• Impulse response:h[n] has finite length

• Transfer function:H(z) has finite number of terms: H(z)=a0z0+a1z-1+a2z-2+...+aMz-M

• Difference equation:y[n]=a0x[n] + a1x[n-1] +a2x[n-2] +...+aMx[n-M]

• Linear phase⇒ group (signal) delay is

constant

• Desired freq. Response:– Hd(ejΩ)

• Practical freq. response:– H(ejΩ)

• Realization: – Mean-square error– Window functions– Implementation of FIR filters

with block diagrammes

• Applications– Discrete time differentiator– Types of filters– Practical examples of filters

Desired Filter and Mean Square Error

( ) ( ) [ ] [ ]

[ ] [ ]

2 2j jd d

1 2 2d d

1E H e H e d h n h n2

h n h n

π ∞Ω Ω

−∞−π

− ∞

−∞ +

= − Ω = −π

∑∫

∑ ∑

Mean square error :

[ ] [ ]

0 for n 0h n h n for 0 n M

0 for M n 0

− ∞ < < ∞

<= ≤ ≤ < <

Desired impulse response

Practical impulse response

Window Functions

[ ] [ ]

( ) ( )( )

Mj jn jM / 2

0 for n 0: w n w n 1 for 0 n M

0 for M n 0

sin (M 1) / 2W e e e

sin / 2Ω Ω − Ω

<= = ≤ ≤ < <

Ω += = − π < Ω ≤ π

Rectangular Window

[ ] [ ] [ ]

( ) ( ) ( ) ( ) ( )

j j j j j( )d d

h n w n h n

1H e W e H e W e H e d2

πΩ Ω Ω Λ Ω−Λ

= ∗ = Λπ ∫

Practical truncated impulse response :

Transfer function :

[ ]Hamming,M

n0.54 0.46cos 2 for 0 n Mw n M

0 for n 0 and n M

− π ≤ ≤ = < >

Hamming window

Example: Rectangular and Hamming Windows

0 0.5 1 1.5 2 2.5 3-70

0 5 10 15 20 250

Rectangular window

0 5 10 15 20 250

Hammingwindow

Rectangular window

Hamming window

Implementation of FIR Filters with Rectangular Window(1)

( ) [ ]Mj2 DTFTj c

e forH e h n0 for

− ΩΩ

Ω ≤ Ω= ←→Ω < Ω ≤ π

Desired frequency response

[ ] ( )c

Mj nj jn 2 c c

1 1 Mh n H e e d

for 0 n <

e d sinc n2 2 2

Ω π − Ω Ω Ω

−π −Ω

Ω Ω = Ω = Ω = − π π π π

∫ ∫

[ ] [ ] [ ]rectc

Mh n h n sinc n and 0 otherwise.

Ω Ω = ⋅ = − π π

Frequency response with rectangular window

Implementation of FIR Filters with Rectangular Window(2)

0 5 10 15 20 25 30 35 40 45 50-0.05

0 5 10 15 20 25 30 35 40 45 500

0 5 10 15 20 25 30 35 40 45 50-0.05

0 1 2 3 4 5 6 7-40

0 1 2 3 4 5 6 7-30

0 1 2 3 4 5 6 7-80

|Wrec|

Implementation of FIR Filters with Hamming Window(1)

( ) [ ]Mj2 DTFTj c

e forH e h n0 for 0

Ω ≤ Ω= ←→< Ω ≤ π

Desired frequency response

[ ]c c

MM j nj jn 2 c c2d

for 0 n< :

1 1 Mh n e e d e d sinc n2 2 2

Ω Ω − Ω− Ω Ω

−Ω −Ω

Ω Ω = Ω = Ω = − π π π π

≤ ∞

∫ ∫

[ ] [ ] [ ]Hammic c

ng,M d

Mh n h n

for 0 n M

nw n 0.54 0.46c sinc n2

and 0 otherwise

− π Ω Ω = ⋅ = ⋅ − π π

Frequency response with Hamming window

Implementation of FIR Filters with Hamming Window(2)

0 5 10 15 20 25 30 35 40 45 50-0.05

0 5 10 15 20 25 30 35 40 45 500

0 5 10 15 20 25 30 35 40 45 50-0.05

0 1 2 3 4 5 6 7-40

0 1 2 3 4 5 6 7-60

0 1 2 3 4 5 6 7-100

|Wham|

FIR Filter Block Diagram Implementation

z-transformed output signal Y(z)= h[0] z0X(z)+ h[1] z-1X(z)+ h[2] z-2X(z)+...+h[n] z-MX(z)

++ Y(z)++

X(z)z-1 z-1 z-1

h[0] h[1] h[2] h[M-1] h[M]

Impulse response: h[n] = [... 0 0 h[0] h[1] h[2] ... h[M] 0 0 ...]

Input signal: x[n]

Output signal: y[n] = h[n]∗x[n] = h[0]x[n] + h[1]x[n-1] + h[2]x[n-2] +...+ h[M]x[n-M]

Discrete Time Differentiatorx yh H←→

( ) ( ) ( ) ( ) ( ) ( ) ( )

[ ] [ ] [ ] ( ) ( ) ( )

( ) sT j T j

z e e e

dy t x t Y s H s X s sX s H s s jdt

y n h n x n Y z H z X z

H z j jTω Ω= = =

= ←→ = = = = ω

= ∗ ←→ =

Ω= ω =

Differentiator in analogue system :

Differentiator in digital system :

Mjj 2d

Mj jn2d

Mth order

Desired transfer function : H e j e

Desired impulse response :

M Mcos n sin n2 2h n j e e d

M Mn s n2 2

− ΩΩ

π− Ω Ω

= Ω − π < Ω ≤ π

π − π − = Ω Ω = − − π −

Normalized digital differentiator filter :

2 for 0 n≤ < ∞

”Windowed” Discrete Time Differentiator

1 for 0 n Mw n

0 for n 0 and for M n 0

M Mcos n sin n2 2h n for 0 n M

M Mn s n2 2

≤ ≤= < < <

π − π − = − ≤ < − π −

Rectangular Window

[ ] [ ] [ ]( ) ( ) ( )

j j jd

Mth orderh n w n h n

H e W e H eΩ Ω Ω

Windowed digital differentiator filter :Windowed impulse response :

Windowed transfer function :

M Mcos n sin nn2 2h n 0.54 0.46cos 2 for 0 n M

M MMn s n2 2

− π ≤ ≤ = < < <

π − π − = − ⋅ − π ≤ < − π −

Hamming Window

Discrete Time DifferentiatorRectangular Window

1 for 0 n Mw n

M Mcos n sin n2 2h n for 0 n M

M Mn s n2 2

≤ ≤= < < <

π − π − = − ≤ < − π −

Rectangular Window

-20 -15 -10 -5 0 5 10 15 20-1

0 1 2 3 4 5 6 70

h[n] |H(ejΩ)|

n Ωπ 2π

Discrete Time DifferentiatorHamming Window[ ]

M Mcos n sin nn2 2h n 0.54 0.46cos 2 for 0 n M

M MMn s n2 2

− π ≤ ≤ = < < <

π − π − = − ⋅ − π ≤ < − π −

Hamming Window

-20 -15 -10 -5 0 5 10 15 20-1

0 1 2 3 4 5 6 70

h[n] |H(ejΩ)|

n Ωπ 2π

Filtering of Speech Signals• Speech must be processed by precise time alignement• FIR-filters with w[n], 0≤n≤M, symmetric or antisymmetric

around M/2 or (M+1)/2 have linear characteristics• Length M+1 of FIR is usually large (fx~100)• Time delay Tsignal=½MTs (group delay)• LP used to remove high frequency noise• LP used to freq.band limit signal • HP used to remove low frequency noise • BP used to limit frequency range, used in speech

processing equipment (fx. telephone 300 < f < 3100Hz)• BS used to remove noise at specific frequencies (fx. 50Hz)

Typical Examples on FIR-filter Characteristics for Filtering of Speech Signals

[ ][ ]

1 2 n1 cos 0 n M ( Hanning or raised cosine window,2 M

0 otherwise with symmetric h n )

M2 ( n )2sin 0 n M (with antisymmetric h[n])

0 otherwise

LP filter h n

BP filter h n

HP filter h n

π − ≤ ≤

− = − =

− = ( )cos n 0 n M (with antisymmetric h[n])

otherwise

g signal s

d dTime delay (= group delay), defined by T Td d

for = T T ½ MT

π ≤ ≤

= − ϕ = − ϕω Ω

= =phase linear FIR filters

LP – FIR filter, An Example

0 1 2 3 4 5 6 70

-20 -15 -10 -5 0 5 10 15 200

[ ]1 2 n1 cos 0 n M2 M

0 otherwiseh n

π − ≤ ≤

h[n] |H(ejΩ)|

BP – FIR filter, An Example

-20 -15 -10 -5 0 5 10 15 20-1

0 1 2 3 4 5 6 70

M2 ( n )2sin 0 n M

0 otherwiseh n

π − ≤ ≤

h[n] |H(ejΩ)|

Signal ProcessingPart 8.6 Digital IIR Filter

Lowpass and Highpass Filters

Magnus Danielsen

IIR-Filter Properties and Applications• Difference equation:

y[n]+a0y[n-1] +... aNy[N] =b0x[n] + b1x[n-1] +...+ bMx[n-M]• Impulse response:

h[n]=h[0], h[1], h[2], ... , h[n], ..... has infinite length• Transfer function:

H(z)=h[0]z0+h[1]z-1+...+h[n]z-n+..... has infinite no. of terms• Causality:

h[n]=0 for n<0 usually applied• Construction of filters by bilinear transform• Butterworth characteristics• Chebyshev characteristics• Elliptic characteristics• LP-filters• HP-filters• BP-filters• BS-filters• Linear distortion – transfer function dependence on frequency

Amplidtude distortionPhase distortion - non-linear phase ⇒ non-constant group (signal) delay

• Equivalizers• Implementation of IIR filters with block diagrammes

Difference Equation for IIR Filters Transfer Function and Infinite Impulse Response

( )1 2 M

0 1 2 M1 2 M

0 1 2 M1 2 n

b b z b z ... b zH za a z a z ... a z

h[0] h[1]z h[2]z .... h[n]z ........

− − −

+ + + +=

+ + + +

= + + + + +

Transfer function :

Difference equation:y[n]+a0y[n-1] +... aNy[N] =b0x[n] + b1x[n-1] +...+ bMx[n-M]

z-transformed difference equation:Y(z)+a0z-1Y(z) +... aNz-NY(z) =b0X(z) + b1z-1X(z) +...+ bMz-MX(z)

[ ] [ ] [ ] [ ] [ ] [ ]h n ....0 0 h 0 h 1 h 2 h 3 ... h n .....= Impulse response :

Bilinear Transform

ssamp s

2 2Sampling time : T

Relation between z and s : z re es jCyclic frequency :Frequency f /(2 )Normalized cyclic frequency : T

π π= =ω ω

= == σ + ω

ω= ω πΩ = ω

Definitions :

Analogue filter characteristic: H(s)Wanted digital filter characteristic: H(z)H(s) and H(z) shall have approximately the same properties

sTe -1 sTs j = tanhe +1 2

= σ + ω→ =

Bilinear transform :z - 1λ = = Σ' + jΩ'z + 1

ω is often

and will bewritten as ωwhich howevercan be confusedwith the stopbandcyclic frequency

Transformation ofs – plane → z – plane → λ - plane

s j= σ + ω ssT' j ' tanh2

λ = Σ + Ω =ssTjz re eΩ= =

Negative s-plane strip: σ<0, -½ωs<ω< ½ωs is immaged in interiour of |z|=1 circle, and negative λ-plane

The frequency axis s=jω interval: σ=0, -½ωs< ω<½ωs is immaged on the |z|=1 circleand the imaginary λ-axis

Positiv s-plane strip: σ>0, -½ωs< ω <½ωs is immaged in exteriour of |z|=1 circle, and positive λ-plane

s-plane

−½ωs

λ-plane

z-plane

Im(z)jz e Ω=

Frequency Axis - Transforms s

j T Ts j ' j ' tanh j tan2 2

T' tan tan tan2 2

ω ω= ω→ λ = Σ + Ω = =

ω ω ΩΩ = = π = ω

The bilinear transformed frequency axis :

( ) ( )( ) ( )( ) ( )

Trigonometric relations :sinh j x j sin x sin j x j sinh x

cosh j x cos x cos j x cosh x

tanh j x j tan x tan j x j tanh x

⋅ = ⋅ ⋅ = ⋅

⋅ = ⋅ =

⋅ = ⋅ ⋅ = ⋅

0Ω’ →∞

IIR LP-filter with Butterworth Response( ) ( )( )

( )ps 0.10.1

log 10 1 / 10 1K

2log /

αα − −≥

Analogue LP-filter:Prototype filter is defined as LP-filter with ωc=1

( ) rK

rr 1 c

1 2r 1H s r 1,2,3,...,K2Ks jexp( j )

−= θ = π =

− θ ω

Digital IIR LP-filter:Prototype filter is defined as LP-filter with Ω’c=1

( ) rK

rr 1 c

1 2r 1H r 1,2,3,...,K2K

jexp( j )'=

−λ = θ = π =

λ− θ Ω

( ) ( )( )( )

ps 0.10.1

log 10 1 / 10 1K

2log ' / '

αα − −≥

1H j ''1'

Ω = Ω+ Ω

ω = ω+ ω

z 1 1 zz 1 1 z

− −λ = =

Transfer Function ofLP IIR-filter with Butterworth Response

( ) rK

rr 1 c

1 2r 1H r 1,2,3,..., K2K

jexp( j )'=

−λ = θ = π =

λ− θ Ω

r1r 1 c

r rr 1 c c

1 2r 1H z r 1,2,3,..., K2K1 1 z jexp( j )

1 1jexp( j ) jexp( j ) z' '

−= θ = π =

−− θ Ω +

− θ − + θ Ω Ω

z 1 1 zz 1 1 z

− −λ = =

+ +c s c

T' tan tan2

ω ω Ω = = π ω

Poles and Zeros Butterworth LP IIR-filters

( ) ( )K1

r rr 1 c c

1 z 2r 1H z r 1, 2,3,..., K2K1 1jexp( j ) jexp( j ) z

+ −= θ = π =

− θ − + θ Ω Ω

1 jexp( j )'z for r 1, 2,..., K1 jexp( j )'

+ θΩ

= =− θ

Poles :

zeroz 1 K-multiple zero= −Zeros :Ω

z-plane

jz e Ω=

5 multpl. zero−⊗

K 5' 0.2=

Example :

5 poles

K=5fc=200 Hzfs=1000 Hz

( ) 15

r1r 1 c

1H z1 1 z jexp( j )' 1 z

− θ Ω + ∏

r2r 1 1 3 5 7 9, , , ,2K 10 10 10 10 10−

θ = π = π π π π π

c s cc

T f' tan tan 0.7272 f

ω Ω = = π =

0 200 400 600 800 1000-100

0 200 400 600 800 1000-4

A(f)=10log|H|

ϕ(f)=∠HFrequency f

Frequency f

Butterworth LP IIR-filter

Block – Diagram IIR Butterworth Filter

( ) ( ) ( )( ) ( )

0.0181 1 z

1 0.509

1 zH z

1 1.2505z53 0.5 7zz 45

− −

−−

-0.5457

1.2505

0.50953

0.0181

( ) ( )

Analogue LP filter: with 0.32491H ss s 1

s 0.03249s 0.10

0.0343s . 24 560 3 9

= + + ω ω

Ts=1.93 sec

LP IIR-filter with Chebyshev Response

1H j1 T

ω = ω+ γ ω

Digital LP-filter:Prototype filter is defined as LP-filter with Ωc=1

( )( )

r rr 1

1 2r 1H C r 1,2,3,...,K2K[ sin j 1 cos ]

−λ = θ = π =

λ + η θ + +η θ∏

1H j ''1 T'

Ω = Ω

+ γ Ω

( )( )

r rr 1

1H s Cs [ sin j 1 cos ]

=+ η θ + + η θ∏

( ) ( )( )( )

ps 0.10.11

cosh 10 1 / 10 1K

cosh /

αα−

− −≥

( ) ( )( )( )

ps 0.10.11

cosh 10 1 / 10 1K

cosh ' / '

αα−

− −≥

Analogue LP-filter:Prototype filter is defined as LP-filter with ωc=1

For Chebyshev filter: Ωc=Ωp

For Chebyshev filter: ωc= ωp

Poles and Zeros of Chebyshev LP IIR-filter

( )1 11 1 1sinh sinh or sinh K sinhK

− − η = = η γ γ

D efinition of param eter η :

2r r r

j cos(sin j ),'

(2r 1)sin j 1 cos where2K

1 sin j 1 cos1z1 1 sin j 1 cos

−λ= − η + θ

= −η θ − + η θ θ = π

− η θ − + η θ+ λ= =

− λ + η θ + + η θ

Poles:

1z 1 is a K-multiple zero1+ λ

λ = ∞ ⇒ = = −− λ

Zeros :

LP Butterworth IIR FiltersT' tan tan

2 2ω Ω

Ω = =

( ) rK

rr 1 c

1 2r 1H r 1,2,3,...,K2K

jexp( j )'=

−λ = θ = π =

λ− θ Ω

( ) ( )( )( )

ps 0.10.1

log 10 1 / 10 1K

2log ' / '

αα − −≥

1H j ''1'

Ω = Ω+ Ω

z 1z 1−

( ) ( )p s

p sc 1 1

0.1 0.12K 2K

' ''10 110 1α α

Ω ΩΩ = =

−−

HP Butterworth IIR Filter:

sT' tan tan2 2ω Ω

Ω = =

( ) rKc

1 2r 1H r 1,2,3,...,K' 2Kjexp( j )

−λ = θ = π =

Ω − θ λ ∏

( ) ( )( )( )

ps 0.10.1

log 10 1 / 10 1K

2 log ' / '

αα − −≥

1H j ''1'

Ω =Ω + Ω

z 1z 1−

( ) ( )p s

1 10.1 0.12K 2K

c p s' ' 10 1 ' 10 1α αΩ = Ω − = Ω −

LP-filtur HP-filtur

' '' ' ' '

Ω Ωλ Ω→ →

Ω λ Ω ΩFrequency transformation :

PolesHP=PolesLP* (i.e. same poles)ZerosHP: λ=0 ⇒ z=1 (K-multiple)

K=5fc=200 Hzfs=1000 Hz

( ) 15

c r1r 1

1H z1 z' jexp( j )1 z

= +Ω − θ −

r2r 1 1 3 5 7 9, , , ,2K 10 10 10 10 10−

c s cc

T f' tan tan 0.7272 f

ω Ω = = π =

Butterworth HP IIR Filter

0 200 400 600 800 1000-4

ϕ(f)=∠H

½fsFrequency f

0 200 400 600 800 1000-100

A(f)=10log|H|

Frequency f½fs

HP Chebyshev IIR Filter:

sT' tan tan2 2ω Ω

Ω = =

1 z1 z

−λ =

+c pFor Chebyshev filters ' 'Ω = Ω

LP-filtur HP-filtur

' '' ' ' '

Ω Ωλ Ω→ →

Ω λ Ω ΩFrequency transformation :

PolesHP=PolesLP* (i.e. same poles)ZerosHP: λ=0 ⇒ z=1 (K-multiple)

( ) ( )( )( )

ps 0.10.11

cosh 10 1 / 10 1K

cosh ' / '

αα−

− −≥

( ) rK2c

r rr 1

1 2r 1H C r 1,2,3,..., K' 2K[ sin j 1 cos ]

−λ = θ = π =

Ω + η θ + + η θ λ ∏

2 2 cK

1H j ''1 T'

Ω =Ω + γ Ω

LP and HP Chebyshev Filters of Order K=5

0 100 200 300 400 500 600 700 800 900 1000-100

0 100 200 300 400 500 600 700 800 900 1000-4

0 100 200 300 400 500 600 700 800 900 1000-100

0 100 200 300 400 500 600 700 800 900 1000-4

LP filter HP filter

fp =200 Hz fs = 1000Hz αp = 3dB γ = 1 η = 0.1775

Frequency (Hz) Frequency (Hz)

Signal ProcessingPart 8.7 Digital IIR Filter

Bandpass, and Bandstop Filters

Magnus Danielsen

Bandpass IIR Filter Frequency Transformation

0 0 0 0

c 0 c 0

' ' ' '' ' ' B' ' ' B' ' '

Ω Ω Ω Ωλ λ Ω Ω→ + → − Ω Ω λ Ω Ω Ω

Frequency transformation :

LP-filtur BP-filtur

p2'Ωp1'Ω s2'Ωs1'Ωs'Ωp'Ω 'Ω 'Ω

1j T j s

Tz 1 1 z z e e ' tan tanz 1 1 z 2 2

−ω Ω

ω− − Ωλ = = = = ⇒ Ω = =

+ +p1 s p2 s

T T' tan ' tan

2 2ω ω

Ω = Ω =

0 s0 p1 p2

T' tan ' '2

ωΩ = = Ω ⋅Ω

s1 s s2 ss1 s2

T T' tan ' tan2 2

ω ωΩ = Ω =

' '' 'H j H j' B ' ' '

Ω ΩΩ Ω→ − Ω Ω Ω

' 'H H' B '

Ω Ωλ λ→ + Ω Ω λ

p2 p1B ' ' '= Ω −Ω

BP Butterworth IIR Filter

( ) ( )( )ps 0.10.1

p1,2s1,2 0 0

0 s1,2 0 p1,2

log 10 1 / 10 1K

'' ' '2 log /' ' ' '

αα − −≥

ΩΩ Ω Ω− − Ω Ω Ω Ω

( ) rK0 0

rr 1 0

1 2r 1H r 1, 2,3,..., K2K' ' jexp( j )

B ' '=

−λ = θ = π =

Ω Ωλ+ − θ Ω λ

1H j '' ''1

B ' ' '

Ω = Ω ΩΩ

+ − Ω Ω

( ) ( ) ( )( ) ( ) ( )

K K1 1

2 2 2K0 0 01 2

r rr 1

1 z 1 zH z

' 2 ' '1 2 1jexp( j ) z z jexp( j )B' B' B' B' B' B'

− −

+ −=

Ω Ω Ω + − θ + − + + + + θ

10 0 0 0

1c 0 c 0

' ' ' '' ' z 1 1 z' B' ' ' B' ' ' z 1 1 z

Ω Ω Ω Ωλ λ Ω Ω − −→ + → − λ = = Ω Ω λ Ω Ω Ω + +

LP-filtur BP-filtur

Bandpass IIR Butterworth Filter of Order K=5

0 100 200 300 400 500 600 700 800 900 1000-100

0 100 200 300 400 500 600 700 800 900 1000-30

-9π/2

-7π/2-8π/2

Frequency f (Hz)

fp1=150 Hz fp2=250 Hzfs=1000 Hz αp=3dB

( ) ( )( )

( )( )

rr 1 0

H z H z

1z' ' jexp( j )

B ' ' z=

= λ =

λΩ Ω+ − θ Ω λ

r2r 1 1 3 5 7 9, , , ,2K 10 10 10 10 10−

1 zz1 z

−λ = λ =

p1 s p2 sp1 p2

0 p1 p2

T T' tan ' tan

2 2B' ' '

ω ωΩ = Ω =

=Ω −Ω

Ω = Ω ⋅Ω

Bandpass IIR Chebyshev I Filter of Order K=5

αp=3dB fp1=150 Hz fp2=250 Hzfs=1000 Hz αp=3dB

( ) ( )( )

( )( )

rr 1 0

H z H z

1z' ' jexp( j )

B ' ' z=

= λ =

λΩ Ω+ − θ Ω λ

r2r 1 1 3 5 7 9, , , ,2K 10 10 10 10 10−

1 zz1 z

−λ = λ =

p1 s p2 sp1 p2

0 p1 p2

T T' tan ' tan

2 2B' ' '

ω ωΩ = Ω =

=Ω −Ω

Ω = Ω ⋅Ω

0 100 200 300 400 500 600 700 800 900 1000-100

0 100 200 300 400 500 600 700 800 900 1000-30

-11π/2

-9π/2-10π/2

Frequency f (Hz)

Bandstop IIR Filter Frequency Transformation

1j T j s

Tz 1 1 z z e e ' tan tanz 1 1 z 2 2

−ω Ω

ω− − Ωλ = = = = ⇒ Ω = =

+ +p1 s p2 s

T T' tan ' tan

2 2ω ω

Ω = Ω = p2 p1B' ' '= Ω −Ωs1 s s2 ss1 s2

T T' tan ' tan2 2

ω ωΩ = Ω =

c 0 0 c 0 0

' 'B' ' B' ' ' ' ' ' ' ' '

− − Ω Ωλ λ Ω Ω

→ + → − Ω Ω Ω λ Ω Ω Ω Ω Frequency transformation :

LP-filtur

s'Ωp'Ω 'Ω

BS-filtur

s2'Ωs1'Ω p2'Ωp1'Ω 'Ω

'' B ' 'H j H j' ' ' '

− ΩΩ Ω → − Ω Ω Ω Ω 1

'B 'H H' ' '

− Ωλ λ → + Ω Ω Ω λ

0 s0 p1 p2

T' tan ' '2

ωΩ = = Ω Ω

Band Stop (BS) Butterworth IIR Filter

( ) ( )( )ps 0.10.1

p1,2 s1,20 0

0 p1,2 0 s1,2

log 10 1 / 10 1K

' '' '2 log /' ' ' '

αα − −≥

Ω ΩΩ Ω− − Ω Ω Ω Ω

( ) r1K0

rr 1 0 0

1 2r 1H r 1, 2,3,..., K2K'B ' jexp( j )

−λ = θ = π =

Ωλ + − θ Ω Ω λ ∏

1H j ''B ' '1

−Ω =

ΩΩ + − Ω Ω Ω

( )( ) ( ) ( )

1 20 0 0K

0 0 002 2 2K

0 0 01 2r r r

1 1 1' 2 ' z ' z' ' ''H z

B' ' ' '1 1 11 jexp( j ) z 2 jexp( j ) z 1 jexp( j )B' B' B' B' B' B'

− −

Ω + + Ω − + Ω + Ω Ω ΩΩ = Ω Ω Ω − + θ + ⋅ ⋅ − θ − + + θ

1 1 10 0

1c 0 0 c 0 0

' 'B' ' B' ' z 1 1 z' ' ' ' ' ' ' z 1 1 z

− − −

Ω Ωλ λ Ω Ω − −→ + → − λ = = Ω Ω Ω λ Ω Ω Ω Ω + +

LP-filtur BS-filtur

Bandstop IIR Butterworth Filter of Order K=5

fp1=150 Hz fp2=250 Hzfsamp=1000 Hz αp=3dB

( ) ( )( )

rr 1 0 0

H z H z

'B ' jexp( j )' '

= λ =

Ωλ + − θ Ω Ω λ ∏

r2r 1 1 3 5 7 9, , , ,2K 10 10 10 10 10−

1 zz1 z

−λ = λ =

s1 s s2 ss1 s2

0 s1 s2

T T' tan ' tan2 2

B' ' '

ω ωΩ = Ω =

=Ω −Ω

Ω = Ω ⋅Ω 0 100 200 300 400 500 600 700 800 900 1000-30

-3π/2

Frequency f (Hz)

an) -5π/2

0 100 200 300 400 500 600 700 800 900 1000-100

Bandstop IIR Chebyshev I Filter of Order K=5

fs1=150 Hz fs2=250 Hzfsamp=1000 Hz αp=3dB

( ) ( )( )

rr 1 0 0

H z H z

'B ' jexp( j )' '

= λ =

Ωλ + − θ Ω Ω λ ∏

r2r 1 1 3 5 7 9, , , ,2K 10 10 10 10 10−

1 zz1 z

−λ = λ =

s1 s s2 ss1 s2

0 s1 s2

T T' tan ' tan2 2

B' ' '

ω ωΩ = Ω =

=Ω −Ω

Ω = Ω ⋅Ω

0 100 200 300 400 500 600 700 800 900 1000-100

0 100 200 300 400 500 600 700 800 900 1000-25

-3π/2

Frequency f (Hz)

an) -π/2

Poles and Zeros IIR Filters of Order K

• LP filter: K complex polesK-multiple zero z = -1

• HP filter: K complex polesK-multiple zero z = 1

• BP filter: 2K complex polesK-multiple zero z = -1K-multiple zero z = +1

• BS filter: 2K complex polesK-multiple complex zero pairs

Resumé: IIR responses of filters of K’th order for LP-, HP-, BP-, and BS-filters

• Same transformations for the LP-, HP-, BP-, and BS- filters are used for Butterworth type response and Chebyshev type response.

• Same transformations are used for other filter types transfer functions as well

Used variables in:

LP filter: λ/Ω’c

HP filter: Ω’c/ λ

BP filter: Ω’0/B’⋅(λ/Ω’0+ Ω’0/ λ) B’=Ω’p2 - Ω’p1= transformed bandwidth of passband

BS filter: B’/Ω’0⋅(λ/Ω’0+ Ω’0/λ)-1 B’=Ω’p2 - Ω’p1=transformed bandwidth of stopband

EqualizersHc(jω) Heq(jω)

( ) ( ) ( ) ( )0j t

FTeq d eq d

eh (t) h t H j H jH j

←→ ω ω =ω

( ) ( ) ( ) ( )s s

FT,d d s

jn T jn T,d d s d s ,eq

h (t) h (t) t nT

H j h nT e h nT e H j

δ=−∞

∞− ω − ω

δ δ=−∞ =

= δ − ←→

ω = = ω

∑ ∑

( ) [ ] ( ) [ ]

[ ] [ ] [ ] ( ) [ ]

DTFT j jnd s d d, d s

DTFT j jneq d d,eq d s

h nT h n H e h n e T

h n w n h n H e h n e T

∞Ω − Ω

δ=−∞

Ω − Ω

= ←→ = Ω = ω

Signal ProcessingPart 9.1: Acoustic Spectrogramme Example

Magnus Danielsen

Speech Production Mechanism

J.R.Deller Jr,J.G.Proakis, J.H.L.Hansen: Discrete-Time Processing of Speech Signals, Prentice Hall, New Jersey, 1987

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-5

Speech:Continuous: f(t)Discrete: f[n]

Window function:Continuous: w(t-t0)Discrete: w[n-n0]

Windowed speech:Continuous: f(t)w(t-t0)Discrete: f[n]·w[n-n0]

Time (s)

[ ] [ ][ ] [ ]

j tFT0

j nFT j j0

j nDFT j j20 kN

Spectrum :Continuous : f (t) w(t t ) F( ) W( )e

Discrete DTFT : f n w n n F(e ) W(e )e

Discrete DFT : f n w n n F(e ) W(e )e

ttn n 0 k N 1T T T

ΩΩ Ω

ΩΩ Ωπ

⋅ − ←→ ω ∗ ω

⋅ − ←→ ∗

Ω= = ω = ≤ ≤ −

Windowing of Speech Signal

Measurement System for Audio and Speech Spectrogrammes

Ampl. Sampl.Freq.= fs

Quant.Micro-phone

Small voltage

Amplified voltage

Sampledvalues

Sampl. data f[n] (numbers)

f[n]·w[n-n0]

LP analogue filter. B = fs/2

Filtred voltage

w[n-n0]Discrete time generator n0

FFTN = nfft

Data loggermemory

F(ejΩ,n0) F·F*

f = fs·Ω /(2π)Ω

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450

3500 Frequency f

Power to color or intensity transformation

Displayt = n0/fs

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-3

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450

Time (s)

Spectrogramme: specgram(mtlb,512,Fs,kaiser(500,5),475)

SignalNumber of samples: 4001Sampling frequency: 7418 HzOverlap: 475Window: Kaiser(500,5)FFT length: 512

0 100 200 300 400 5000

Spectrum

0 0.05 0.1 0.15 0.2 0.25 ... ...

No overlap

0 0.1 0.2 0.3 0.4 0.50

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450

specgram(mtlb,512,Fs,kaiser(500,5),0) specgram(mtlb,512,Fs,kaiser(500,5),475)

0 0.05 0.1 0.15 0.2 0.25 ... ...

Overlap: 475

FFT length = 515, Kaiser window, window lenth = 500, β=5

Influence of Overlap

Influence of Length of Window

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450

specgram(mtlb,512,Fs,kaiser(500,5),475)Window length = 500

~ 0.0634 secFFT length = 512

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

specgram(mtlb,1024,Fs,kaiser(1000,5),975)

Window length = 1000~ 0.1287 sec

FFT length = 1024

0 0.1 0.2 0.3 0.4 0.50

Window length = 200 ~ 0.02696 sec

FFT length = 512

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450

Influence of Window Shape

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450

⇒-100 0 100 200 300 400 500 6000

1.4Rect.window = kaiser(500,0)

-100 0 100 200 300 400 500 6000

1.4Kaiser.window = kaiser(500,5)

Window length=500

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450

0 100 200 300 400 5000

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450

specgram(mtlb,512,Fs,hamming(500),475)

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450

specgram(mtlb,512,Fs,hanning(500),475)

0 100 200 300 400 5000

Influence of Window Shape

Signal ProcessingPart 9.2: Delays in Systems

Magnus Danielsen

Delay in Transmission Systems

SystemNo magnitude distortionTransfer function H(ω)

xphys(t) yphys(t)

Definition of system:•Type of system, e.g.

•Communication system•Measurement system•Audio Hi-Fi system•Television system

•Input signal: xphys(t)•Output signal: yphys(t)•Group delay = signal delay: Tgr•Phase delay: Tph•No magnitude distortion in the transmission band

Group Delay and Phase Delay

-3 -2 -1 0 1 2 3-1

Time (arb.units)

xphys(t)

x0(t)x0(t-Tgr)

yphys(t)Sig

( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( )

( ) ( ) ( )

phys 0 0

0 0 0 0

x t x t cos j t

½x t exp j t ½x t exp j t

Re x t exp j t

x t x t exp j t

= ω + − ω

Input signal :

( ) ( ) ( )phys 0 gr 0 phy t x t T cos j (t T )= − ω −

Output signal = delayed input signal :

Definition of System Transfer Function and Input Signal

( ) ( )( ) ( ) ( ) ( )

FT0 0 0 0

x t x t exp j t X

←→ ω

= ω ←→ ω−ω

Envelope :

Input signal :

( ) ( )( )

( ) ( ) ( ) ( )0

0 0 0 0 0

with no magnitude distortion :

H H exp j H constant

gives the linear frequency variation :

dω=ω

ω = Ψ ω =

Ψ ωΨ ω = Ψ + ω− ω = Ψ + Ψ ⋅ ω− ω

Transfer function

Phase - first order Taylor approx. around ω

Definition of Group Delay and Phase Delay

( ) ( ) ( )( )( )( )( )

( )( ) ( )

( ) ( )( ) ( )

0 0 0 0

0 0 0 0 0

Y H j X( j ) H exp j X ( )

H exp j ' X ( )

X ( ) exp j '

x (t ') exp j (t '

x (t ')

H exp j '

H exp j

Ψ −Ψ ⋅ω

ω = ω ⋅ ω = Ψ ω ⋅ ω− ω

= Ψ + Ψ ⋅ ω− ω ⋅ ω− ω

ω− ω ⋅ Ψ ⋅ω

+Ψ − Ψ ⋅ω Ψ ⋅ ω

+ Ψ ω⋅ Ψ= ⋅

Output signal Fourier transform :

Output signal :

( )( ) ( )phy

000 0s 0 0

x (t ')y cos (H )/t

t+ Ψ ω + Ψ⋅ ⋅ ω=

T Ψ= −

Phase delay :

dT 'd ω=ω

Ψ= −Ψ = −

Group delay

H(jω) X(jω) Y(jω)

Phase and Group Delay Based on Transfer Function

( ) ( )( )( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )

H j H exp j

ln H j ln H j j

½ ln H j ½ ln H j j

ω = Ψ ω

ω = ω + Ψ ω = ω + ω + Ψ ω

= ω + − ω + Ψ ω

Transfer function in Fourier transform formulation :

( ) ( )( )

( )( )gr

H jj ln2 H j

d H jdT ½ lnd d j H j

ωΨ ω = − − ω

Ψ ω ωω = − = − ω ω − ω

Phase :

Group delay :

Spectrum of input pulse

0 0.5 1 1.5 2

x 10-3

X Axis

X Y P lot

Time (sec)

0 0.5 1 1.5 2

x 10-3

X Axis

X Y P lot

Time (sec)t1=0.5ms

t2=0.757 ms

Group delay=t2 - t1=0.257ms

XY Graph1

XY Graphs -3*10^4s+10^102

s +3*10^4s+10^102

Transfer Fcn2

s -3*10^4s+10^102

s +3*10^4s+10^102

Transfer Fcn1

Sine Wave1

Sine Wave - envelope

PulseGenerator

Product1Product

BufferedFFT Scope

4th order all-pass filter

Simulation of Group Delay of Pulse in4th Order All-pass Filter by Simulink

Group delay for ω=105 rad/sec:

(4π)/ ∆ω = 0.257 ms

( )2 4

2s 3 10 s 1H ss 3 10 s 1

− ⋅ += + ⋅ +

Transfer function4 5

Double complex pole-pairs: s 1.5 10 j 10Double complex zero-pairs: s 1.5 10 j 10

≅ − ⋅ ± ⋅

≅ + ⋅ ± ⋅

Poles and zeros

3 104 Arctan1

− ⋅ ωϕ = − ⋅ − ω

Phase :

4th Order All-pass Filter

0 0.5 1 1.5 2 2.5 3 3.5

Cyclic frequency (rad/sec)

∆ω=0.489·105 rad/sec

Phase delay for ω=105 rad/sec:

(2π)/ω = 0.0628 ms

( ) ( ):

y t (1 m x(t)) cos t= + ⋅ ⋅ ω

Amplitude modulated signals -3*10^4s+10^102

s +3*10^4s+10^102

Transfer Fcn2

s -3*10^4s+10^102

s +3*10^4s+10^102

Transfer Fcn1Sum2

Sine Wave1

Sine Wave - envelope1

Scope1

Product1 Mux

Constant21

Constant1

BufferedFFT Scope

( )x(t)

cos t=

− ω

y(t) gry(t T )−

Delay in AM-system Simulink Simulation

0 5 10 15

Frequency (rad/sec)

Spectrum of AM-modulated signal

Tgr=0.25 ms

Time s

gry(t T )−

Signal ProcessingPart 9.3: Seismic Example

Magnus Danielsen

Seismic Ship and Streamer Configuration

B K Galloway and S L Klemperer (Dept.Geoph. Stanford Univ., CA); J R Childs (Menlo Park, CA); Bering-Chukchi Working Group (Inst.of the Lithosphere and Inst.of Oceanology, Russian Academy of Sciences, Western Washington University, Lamont-Doherty Earth Observatory, Columbia University)

Wing-kei Yu, Rice University, Houston, Texas

SeismicSource -Airgun

Hydrophone cableHydrophones

Convolutional Model

Source Receiver

Sea surface

Seismic wavelet: h[k] = s[k]∗f[k]Source signal: s0[k]

”Source” signature:s[k]=s0[k]∗p[k]

Reverberationfunction: f[k]

Received signal: y[k] = h[k]∗ε[k] + n[k] = s0[k]∗p[k]∗f[k]∗ε[k] + n[k]

Noise: n[k]

Received signal: y[k]

Sea floor

Reflectivity function: ε[k]

Undergroundhorizons

Seismic Source - Airgun

A. Ziolkowski, J. R. Underhill, R.G.K. Johnston Wavelets, well ties, and the searchfor subtle stratigraphic trapsGEOPHYSICS, VOL. 63, 1998; P. 297–313

Maximum and Minimum Phase Example on Discrete Signal

1 11 z 1 z3 4

− −

+= + +

Minimum phase :2 zH

1 11 z 1 z3 4

− −

+= + +

-4 -2 0 2 4-4

4Maximum phase

-4 -2 0 2 4-4

4Minimum phase

-2 0 2 4 6 8 1 0-1 .5

1M a xim um p ha se

-2 0 2 4 6 8 1 0-0 .5

2M inim um p ha se

[ ] [ ] [ ]n n

Maximum phase impulse response :

1 1h n 20 u n 21 u n3 4

= − +

[ ] [ ] [ ]n n

Minimum phase impulse response :

1 1h n 4 u n 6 u n3 4

= − +

All-pass Filter for Converting Max-phase to Min-phase – An Example

max1 1

1 11 z 1 z3 4

− −

+= + +

min1 1

Minimum phase :2 zH

1 11 z 1 z3 4

− −

+= + +

- 4 - 2 0 2 4- 4

4M a x i m u m p h a s e

c y c li c fr e q u e n c y ( r a d /s e c )

- 4 - 2 0 2 4- 4

4M i n i m u m p h a s e

c y c li c fr e q u e n c y ( r a d /s e c )

-2 0 2 4 6 8 10-1 .5

1M aximum phase

-2 0 2 4 6 8 10-0 .5

2M inimum phase

Phase: Impulse response

( ) ( )( ) ( )

1min j

All pass filter transfer function :H z 2 zH z H e 1H z 1 2z

+= = =

Properties of FunctionsReceived signal: y[k] = h[k]∗ε[k] = s0[k]∗p[k]∗f[k]∗ε[k] + n[k]

Source signal: s0[k] Minimum phase signal

Source signature: s[k] = s0[k]∗p[k] Mixed phase signal

Spike filter response: p[k] All pass response

Reverberation function: f[k] Results from multiple reflections between bottom and surface

Reflectivity function: ε[k] Results from object to be measured

Reflectivity autocorr.fct.: Rεε[k]=σ2δ[k] ε as a random-like property

Noise: n[k] Can be random or coherent

Signature Deconvolution (1)

Source Receiver

Sea surface

Seismic wavelet: h[k] = s[k]∗f[k]Source signature: s[k]=s0[k]∗p[k]Reverberation function: f[k]

DTFT j

DTFT-1

Source signal s [k] is a

p[k] P e is an

and can be compensated for by convolution with an

inverse filter with the response q[k] = p [k]

Ω←→

← →

minimum phase signal

all - pass response function

( )( ) ( )

[ ] ( ) ( )[ ]

2j j ( ) j

Properties: P e e P e =1 (Constant energy spectrum)

1 q[k] p[k]= k Q eP e

p k is stable and causal q[k] is anticausal

Ω θ Ω Ω

∗ δ =

Signature Deconvolution (2)

Received signal: y[k] = h[k]∗ε[k] = s0[k]∗p[k]∗f[k]∗ε[k] + n[k]

Noise will be ignored (this is not always possible):n[k]∼0

Inverse source signature must be found: s-1[k]=s0

-1∗p-1[k]

Processing is now performed finding:s-1[k] ∗y[k] = Σn s-1[n] ∗y[n-k] =f[k]∗ε[k]

Reflection and Transmission Coefficients from Horizons with Perpendicular Incidents

Medium 1

Medium 2

Density of mass: ρ1Velocity of sound: ν1Impedance: Z1=ρ1ν1

Density of mass: ρ2Velocity of sound: ν2Impedance: Z2=ρ2ν2

Reflecting surface

21 1 22

Z Z: cZ Z

2Z: t 1 cZ Z

Z 4Z Z: tZ Z Z

−= +

= + =+

Reflection coefficient

Power reflection factor

Transmission coefficient

Power transmission factor

Reverberation Deconvolution (1) (deterministic)Seismic trace after signature deconvolution:u[k] = f[k]∗ε[k]

Reverberation function: f[k] Reflectivity function: ε[k]

Sea surface

Sea floor

1 -c (-c)2 (-c)3 (-c)3 (-c)3 .....

Down going signal: g[k] = 1 0 0 .... 0 -c 0 0 .... (-c)2 0 0 ... 0 (-c)3 0 ...k = 0 ................ N ............. 2N ................. 3N .......

Reverberation Deconvolution (2) (deterministic)

( ) ( ) ( )

2 3 4N 2 N 3N 4 NN

1G(z) 1 cz c z c z c z .....

1 czInverse z transform : G (z) 1 czInverse impulse response : g k 1 0 0 0...0 c 0.........

k 0 1 2 3...... N ...........

− − − −−

− −

= − + − + − + − =+

− = +

z - transform of downgoing signal :

Down going signal: g[k] = 1 0 0 .... 0 -c 0 0 .... (-c)2 0 0 ... 0 (-c)3 0 ...k = 0 ................ N ............. 2N ................. 3N .......

( )( )

( )[ ]

2 N 2 2 N 3 3N 4 4 N2N

21 N N 2 2 N

1F(z) G(z) 1 2cz 3c z 4c z 5c z .....

Inverse z transform : F (z) 1 cz 1 2cz c z

Inverse impulse response : f k 1 0 0 0.

− − − −

= = = − + − ++

− = + = + +

z - transform of two - way transmission of signal :

FIR filter

FIR filter 2..0 2c 0...0 c 0 0 0 0........k 0 1 2 3...... N .........2N....................=

Reverberation Deconvolution (3)

Sea surface

Sea floor

Target horizon

Direct signalSea surface

Sea floor

Target horizon

1st order multiple signal

Sea surface

Sea floor

Target horizon

2nd order multiple signal

N 2 2 N 3 3N

F(z)1 2cz 3c z 4c z ...− − −

− + − +

z - transform of two - waytransmission of signal :

Signal Processing - An Introduction Processing – An Introduction ... “Analog and Digital Signal Processing”, ... – Biological: EEG interfered by ECG – Signal analytical:

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